title
Maximum Likelihood For the Normal Distribution, step-by-step!!!

description
Calculating the maximum likelihood estimates for the normal distribution shows you why we use the mean and standard deviation define the shape of the curve. NOTE: This is another follow up to the StatQuests on Probability vs Likelihood https://youtu.be/pYxNSUDSFH4 and Maximum Likelihood: https://youtu.be/XepXtl9YKwc Viewers asked for worked out examples, and this one is super mathy, but I just couldn't say "no"! For a complete index of all the StatQuest videos, check out: https://statquest.org/video-index/ If you'd like to support StatQuest, please consider... Buying The StatQuest Illustrated Guide to Machine Learning!!! PDF - https://statquest.gumroad.com/l/wvtmc Paperback - https://www.amazon.com/dp/B09ZCKR4H6 Kindle eBook - https://www.amazon.com/dp/B09ZG79HXC Patreon: https://www.patreon.com/statquest ...or... YouTube Membership: https://www.youtube.com/channel/UCtYLUTtgS3k1Fg4y5tAhLbw/join ...a cool StatQuest t-shirt or sweatshirt: https://shop.spreadshirt.com/statquest-with-josh-starmer/ ...buying one or two of my songs (or go large and get a whole album!) https://joshuastarmer.bandcamp.com/ ...or just donating to StatQuest! https://www.paypal.me/statquest Lastly, if you want to keep up with me as I research and create new StatQuests, follow me on twitter: https://twitter.com/joshuastarmer 0:00 Awesome song and introduction 0:45 Overview of the normal distribution equation 1:41 Example with one data point 5:38 Example with two data points 7:35 Example in 'n' data points 8:08 Solving for the MLEs for mu and sigma 18:54 Review of concepts Correction: 2:39 I said likelihood=0.03 for mu=30, but mu=28 is in the equation. #statquest #MLE #statistics

detail
{'title': 'Maximum Likelihood For the Normal Distribution, step-by-step!!!', 'heatmap': [{'end': 121.131, 'start': 90.972, 'weight': 1}], 'summary': 'Explains maximum likelihood estimation for the normal distribution, covering parameters mu and sigma, likelihood calculation, and finding optimal estimates with examples, resulting in likelihood values of 0.12 and a step-by-step simplification of the equation.', 'chapters': [{'end': 343.756, 'segs': [{'end': 212.039, 'src': 'heatmap', 'start': 90.972, 'weight': 0, 'content': [{'end': 95.356, 'text': 'And a smaller value for sigma makes the normal curve taller and narrower.', 'start': 90.972, 'duration': 4.384}, {'end': 106.567, 'text': "In this stat quest, we're going to use the likelihood of the normal distribution to find the optimal parameters for mu, the mean, and sigma,", 'start': 96.617, 'duration': 9.95}, {'end': 109.91, 'text': 'the standard deviation given some data x.', 'start': 106.567, 'duration': 3.343}, {'end': 115.769, 'text': "Let's start with the simplest data set of all, a single measurement.", 'start': 111.448, 'duration': 4.321}, {'end': 121.131, 'text': "Here, we've measured a single mouse, and it weighs 32 grams.", 'start': 116.95, 'duration': 4.181}, {'end': 132.075, 'text': 'Now, just to see what happens, we can overlay a normal distribution with mu equals 28 and sigma equals 2 onto the data.', 'start': 122.532, 'duration': 9.543}, {'end': 140.238, 'text': 'To determine the likelihood of the data given this curve, we can plug the numbers into the likelihood function.', 'start': 133.896, 'duration': 6.342}, {'end': 143.752, 'text': 'and then plug the numbers into this equation.', 'start': 141.311, 'duration': 2.441}, {'end': 152.857, 'text': "And here's the equation with mu equals 28, sigma equals 2, and x equals 32 plugged in.", 'start': 145.013, 'duration': 7.844}, {'end': 155.639, 'text': 'Now just do the math.', 'start': 154.298, 'duration': 1.341}, {'end': 158.02, 'text': 'Plug and chug, plug and chug.', 'start': 156.54, 'duration': 1.48}, {'end': 168.939, 'text': 'And the likelihood of the curve with mu equals 30 and sigma equals 2, given the data, is 0.03.', 'start': 159.321, 'duration': 9.618}, {'end': 174.743, 'text': 'Thus, the y-axis value here is 0.03.', 'start': 168.939, 'duration': 5.804}, {'end': 181.808, 'text': 'Now we can shift the distribution a little bit to the right by setting mu equals 30 and then calculate the likelihood.', 'start': 174.743, 'duration': 7.065}, {'end': 186.571, 'text': 'Again, we just plug the numbers into the likelihood function.', 'start': 183.229, 'duration': 3.342}, {'end': 189.574, 'text': 'And now we just plug and chug.', 'start': 187.832, 'duration': 1.742}, {'end': 200.37, 'text': 'And the likelihood of the new curve with mu equals 30 and sigma equals 2, given the data, is 0.12.', 'start': 191.035, 'duration': 9.335}, {'end': 205.854, 'text': 'Thus, the y-axis value here is 0.12.', 'start': 200.37, 'duration': 5.484}, {'end': 212.039, 'text': 'If we decide to fix sigma equals 2 so that it is a given, just like the data,', 'start': 205.854, 'duration': 6.185}], 'summary': 'Using likelihood of normal distribution to optimize mu and sigma, with mu=30, sigma=2, and likelihood of 0.12.', 'duration': 155.236, 'max_score': 90.972, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/Dn6b9fCIUpM/pics/Dn6b9fCIUpM90972.jpg'}, {'end': 267.319, 'src': 'embed', 'start': 237.975, 'weight': 3, 'content': [{'end': 245.139, 'text': 'The y-axis is the likelihood value, and the x-axis is for the different values that we plug in for mu.', 'start': 237.975, 'duration': 7.164}, {'end': 250.842, 'text': 'Each time we change mu, we calculate the likelihood and plot it.', 'start': 246.7, 'duration': 4.142}, {'end': 259.406, 'text': 'We can identify the peak in the likelihood graph by determining where the slope of the curve equals zero.', 'start': 253.223, 'duration': 6.183}, {'end': 267.319, 'text': 'In this case, the slope equals 0 when mu equals 32.', 'start': 261.058, 'duration': 6.261}], 'summary': 'Identified peak likelihood at mu=32', 'duration': 29.344, 'max_score': 237.975, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/Dn6b9fCIUpM/pics/Dn6b9fCIUpM237975.jpg'}, {'end': 343.756, 'src': 'embed', 'start': 291.817, 'weight': 4, 'content': [{'end': 299.383, 'text': 'That said, if we had more data, then we could plot the likelihoods for different values of sigma,', 'start': 291.817, 'duration': 7.566}, {'end': 305.748, 'text': 'and the maximum likelihood estimate for sigma would be at the peak, where the slope of the curve equals zero.', 'start': 299.383, 'duration': 6.365}, {'end': 315.696, 'text': 'The goal of this super simple example is to convey the basic concepts of how to find the maximum likelihood estimates for mu and sigma.', 'start': 307.329, 'duration': 8.367}, {'end': 320.365, 'text': 'To solve for the maximum likelihood estimate for mu.', 'start': 316.923, 'duration': 3.442}, {'end': 326.508, 'text': "we treat sigma like it's a constant and then find where the slope of its likelihood function is zero.", 'start': 320.365, 'duration': 6.143}, {'end': 331.21, 'text': 'And to solve for the maximum likelihood estimate for sigma.', 'start': 327.788, 'duration': 3.422}, {'end': 336.753, 'text': "we treat mu like it's a constant and then find where the slope of its likelihood function is zero.", 'start': 331.21, 'duration': 5.543}, {'end': 343.756, 'text': "The example with one measurement kept the math simple, but now I think we're ready to dive in a little deeper.", 'start': 338.233, 'duration': 5.523}], 'summary': 'Explaining how to find maximum likelihood estimates for mu and sigma with simple example and readiness for deeper dive.', 'duration': 51.939, 'max_score': 291.817, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/Dn6b9fCIUpM/pics/Dn6b9fCIUpM291817.jpg'}], 'start': 0.129, 'title': 'Maximum likelihood estimation for normal distribution', 'summary': 'Covers maximum likelihood estimation for the normal distribution, focusing on parameters mu and sigma, likelihood calculation, and finding optimal estimates, with an example resulting in a likelihood of 0.12 for a specific curve and a simple illustration for identifying peak in the likelihood graph.', 'chapters': [{'end': 115.769, 'start': 0.129, 'title': 'Maximum likelihood for normal distribution', 'summary': 'Explains maximum likelihood for the normal distribution, highlighting the parameters mu and sigma, and how it is used to find optimal parameters given data, starting with a single measurement.', 'duration': 115.64, 'highlights': ['The likelihood of the normal distribution is used to find the optimal parameters for mu, the mean, and sigma, the standard deviation, given some data x, starting with a single measurement.', "The Greek character mu determines the location of the normal distribution's mean. A smaller value for mu moves the mean to the left, and a larger value for mu moves the mean to the right.", "The Greek character sigma is the standard deviation and determines the normal distribution's width. A larger value for sigma makes the normal curve shorter and wider, while a smaller value makes it taller and narrower."]}, {'end': 212.039, 'start': 116.95, 'title': 'Likelihood calculation with normal distribution', 'summary': 'Introduces the process of calculating the likelihood of data given a normal distribution, with an example of a mouse weighing 32 grams and the calculation resulting in a likelihood of 0.12 for a specific curve.', 'duration': 95.089, 'highlights': ['The likelihood of the new curve with mu equals 30 and sigma equals 2, given the data, is 0.12, demonstrating the process of calculating likelihood with specific values. (Quantifiable data: likelihood value of 0.12)', 'The likelihood of the curve with mu equals 30 and sigma equals 2, given the data, is 0.03, indicating the initial likelihood calculation process with specific values. (Quantifiable data: likelihood value of 0.03)', 'The mouse weighs 32 grams, serving as the specific data used for the likelihood calculation example. (Quantifiable data: weight of the mouse is 32 grams)']}, {'end': 343.756, 'start': 212.039, 'title': 'Finding maximum likelihood estimates', 'summary': 'Explains how to find the maximum likelihood estimates for mu and sigma using a simple example, illustrating how to identify the peak in the likelihood graph and solve for the maximum likelihood estimates.', 'duration': 131.717, 'highlights': ['The maximum likelihood estimate for mu is found by identifying the peak in the likelihood graph, where the slope of the curve equals zero, which in this case is at mu equals 32.', 'To find the maximum likelihood estimate for sigma, one needs more than one measurement and plots the likelihoods for different values of sigma, with the maximum likelihood estimate at the peak where the slope of the curve equals zero.', 'The process for finding the maximum likelihood estimate for mu involves treating sigma like a constant and finding where the slope of its likelihood function is zero, while for finding the maximum likelihood estimate for sigma, mu is treated like a constant and the slope of its likelihood function is zero.']}], 'duration': 343.627, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/Dn6b9fCIUpM/pics/Dn6b9fCIUpM129.jpg', 'highlights': ['The likelihood of the new curve with mu equals 30 and sigma equals 2, given the data, is 0.12, demonstrating the process of calculating likelihood with specific values. (Quantifiable data: likelihood value of 0.12)', 'The likelihood of the curve with mu equals 30 and sigma equals 2, given the data, is 0.03, indicating the initial likelihood calculation process with specific values. (Quantifiable data: likelihood value of 0.03)', 'The mouse weighs 32 grams, serving as the specific data used for the likelihood calculation example. (Quantifiable data: weight of the mouse is 32 grams)', 'The maximum likelihood estimate for mu is found by identifying the peak in the likelihood graph, where the slope of the curve equals zero, which in this case is at mu equals 32', 'To find the maximum likelihood estimate for sigma, one needs more than one measurement and plots the likelihoods for different values of sigma, with the maximum likelihood estimate at the peak where the slope of the curve equals zero', 'The process for finding the maximum likelihood estimate for mu involves treating sigma like a constant and finding where the slope of its likelihood function is zero, while for finding the maximum likelihood estimate for sigma, mu is treated like a constant and the slope of its likelihood function is zero', 'The likelihood of the normal distribution is used to find the optimal parameters for mu, the mean, and sigma, the standard deviation, given some data x, starting with a single measurement', "The Greek character mu determines the location of the normal distribution's mean. A smaller value for mu moves the mean to the left, and a larger value for mu moves the mean to the right", "The Greek character sigma is the standard deviation and determines the normal distribution's width. A larger value for sigma makes the normal curve shorter and wider, while a smaller value makes it taller and narrower"]}, {'end': 742.545, 'segs': [{'end': 405.337, 'src': 'embed', 'start': 344.966, 'weight': 3, 'content': [{'end': 349.59, 'text': "So let's use a two-sample data set to calculate the likelihood of a normal distribution.", 'start': 344.966, 'duration': 4.624}, {'end': 351.191, 'text': 'Crazy times.', 'start': 350.19, 'duration': 1.001}, {'end': 364.502, 'text': "To keep track of things, let's call the mouse that weighs 32 grams x sub 1 and the mouse that weighs 34 grams x sub 2.", 'start': 352.592, 'duration': 11.91}, {'end': 373.389, 'text': "And again, just to see what happens, let's overlay a normal distribution with mu equals 28 and sigma equals 2 onto the data.", 'start': 364.502, 'duration': 8.887}, {'end': 381.905, 'text': "We've already seen how to calculate the likelihood for this curve given x sub 1, the mouse that weighs 32 grams.", 'start': 375.08, 'duration': 6.825}, {'end': 390.47, 'text': 'And we can calculate the likelihood for the curve given x sub 2 by plugging in 34 into this likelihood function.', 'start': 383.406, 'duration': 7.064}, {'end': 399.497, 'text': "But what's the likelihood of this normal curve given both x sub 1 and x sub 2??", 'start': 392.192, 'duration': 7.305}, {'end': 401.298, 'text': "In other words, what's this??", 'start': 399.497, 'duration': 1.801}, {'end': 405.337, 'text': 'Because these measurements are independent.', 'start': 402.677, 'duration': 2.66}], 'summary': 'Using a two-sample data set to calculate likelihood of normal distribution.', 'duration': 60.371, 'max_score': 344.966, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/Dn6b9fCIUpM/pics/Dn6b9fCIUpM344966.jpg'}, {'end': 577.076, 'src': 'embed', 'start': 497.254, 'weight': 0, 'content': [{'end': 501.935, 'text': "Here's the likelihood function without any value specified for mu and sigma.", 'start': 497.254, 'duration': 4.681}, {'end': 507.797, 'text': 'It equals the product of the likelihood functions for the n individual measurements.', 'start': 503.096, 'duration': 4.701}, {'end': 511.078, 'text': "And here's what the equation looks like.", 'start': 509.297, 'duration': 1.781}, {'end': 516.239, 'text': 'What we need to do is take two different derivatives of this equation.', 'start': 512.097, 'duration': 4.142}, {'end': 523.253, 'text': "One derivative will be with respect to mu, when we treat sigma like it's a constant.", 'start': 517.811, 'duration': 5.442}, {'end': 530.416, 'text': 'And we can find the maximum likelihood estimate for mu by finding where this derivative equals zero.', 'start': 524.694, 'duration': 5.722}, {'end': 537.358, 'text': "The other derivative will be with respect to sigma when we treat mu like it's a constant.", 'start': 531.936, 'duration': 5.422}, {'end': 544.521, 'text': 'And we can find the maximum likelihood estimate for sigma by finding where this derivative equals zero.', 'start': 538.899, 'duration': 5.622}, {'end': 551.011, 'text': "But before we try to take any derivatives, let's take the log of the likelihood function.", 'start': 546.189, 'duration': 4.822}, {'end': 556.914, 'text': 'We do this because it makes taking the derivative way, way easier.', 'start': 552.212, 'duration': 4.702}, {'end': 565.037, 'text': 'And the likelihood function and the log of the likelihood function both peak at the same values for mu and sigma.', 'start': 558.114, 'duration': 6.923}, {'end': 572.16, 'text': "I know it's hard to tell that the log likelihood function for sigma peaks here, but just take my word for it.", 'start': 566.458, 'duration': 5.702}, {'end': 573.321, 'text': 'I double checked.', 'start': 572.521, 'duration': 0.8}, {'end': 577.076, 'text': "Here's the log of the likelihood.", 'start': 575.435, 'duration': 1.641}], 'summary': 'Finding maximum likelihood estimates for mu and sigma using derivatives and log likelihood function.', 'duration': 79.822, 'max_score': 497.254, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/Dn6b9fCIUpM/pics/Dn6b9fCIUpM497254.jpg'}], 'start': 344.966, 'title': 'Normal distribution likelihood and estimation', 'summary': 'Covers the calculation of likelihood for a normal distribution using a two-sample dataset and explains maximum likelihood estimation for mu and sigma, with an example of mice weighing 32 grams and 34 grams, resulting in a small likelihood value. it also simplifies the equation step by step.', 'chapters': [{'end': 441.428, 'start': 344.966, 'title': 'Normal distribution likelihood calculation', 'summary': 'Discusses the calculation of the likelihood of a normal distribution using a two-sample dataset, with a specific example of mice weighing 32 grams and 34 grams, resulting in a small likelihood value.', 'duration': 96.462, 'highlights': ['The likelihood of a normal curve given both x sub 1 and x sub 2 is calculated by multiplying the likelihood of the distribution given x sub 1 with the likelihood of the distribution given x sub 2, resulting in a small number.', 'Using a two-sample dataset, the likelihood of a normal distribution with specific parameters mu equals 28 and sigma equals 2 was calculated, with the example of mice weighing 32 grams and 34 grams.']}, {'end': 742.545, 'start': 443.049, 'title': 'Maximum likelihood estimation for normal distribution', 'summary': 'Explains how to calculate the likelihood of a normal distribution with multiple measurements, deriving the maximum likelihood estimates for mu and sigma and simplifying the equation step by step.', 'duration': 299.496, 'highlights': ['The likelihood function equals the product of the likelihood functions for the n individual measurements, with n data points, and we multiply together all n individual likelihood functions.', 'Deriving the maximum likelihood estimate for mu by finding where the derivative equals zero with respect to mu, treating sigma as a constant, and doing the same for sigma by finding where the derivative equals zero with respect to sigma, treating mu as a constant.', 'Taking the log of the likelihood function makes taking the derivative easier, and both the likelihood function and the log of the likelihood function peak at the same values for mu and sigma.']}], 'duration': 397.579, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/Dn6b9fCIUpM/pics/Dn6b9fCIUpM344966.jpg', 'highlights': ['The likelihood function equals the product of the likelihood functions for the n individual measurements, with n data points, and we multiply together all n individual likelihood functions.', 'Deriving the maximum likelihood estimate for mu by finding where the derivative equals zero with respect to mu, treating sigma as a constant, and doing the same for sigma by finding where the derivative equals zero with respect to sigma, treating mu as a constant.', 'Taking the log of the likelihood function makes taking the derivative easier, and both the likelihood function and the log of the likelihood function peak at the same values for mu and sigma.', 'The likelihood of a normal curve given both x sub 1 and x sub 2 is calculated by multiplying the likelihood of the distribution given x sub 1 with the likelihood of the distribution given x sub 2, resulting in a small number.', 'Using a two-sample dataset, the likelihood of a normal distribution with specific parameters mu equals 28 and sigma equals 2 was calculated, with the example of mice weighing 32 grams and 34 grams.']}, {'end': 1182.593, 'segs': [{'end': 810.382, 'src': 'embed', 'start': 779.653, 'weight': 3, 'content': [{'end': 782.675, 'text': "We'll start by taking the derivative with respect to mu.", 'start': 779.653, 'duration': 3.022}, {'end': 793.022, 'text': "This derivative is the slope function for the log of the likelihood curve, and we'll use it to find the peak, aka where the slope equals zero.", 'start': 783.756, 'duration': 9.266}, {'end': 799.086, 'text': "The first term doesn't contain mu, so its derivative is zero.", 'start': 794.863, 'duration': 4.223}, {'end': 805.09, 'text': "The second term doesn't contain mu either, so its derivative is also zero.", 'start': 800.207, 'duration': 4.883}, {'end': 810.382, 'text': 'The third term contains mu, so now we have to work.', 'start': 806.96, 'duration': 3.422}], 'summary': 'Derivative with respect to mu used to find peak in likelihood curve.', 'duration': 30.729, 'max_score': 779.653, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/Dn6b9fCIUpM/pics/Dn6b9fCIUpM779653.jpg'}, {'end': 914.88, 'src': 'embed', 'start': 848.126, 'weight': 0, 'content': [{'end': 849.726, 'text': "And the 2's cancel out.", 'start': 848.126, 'duration': 1.6}, {'end': 855.008, 'text': 'And that means the derivative of this is this.', 'start': 851.327, 'duration': 3.681}, {'end': 872.098, 'text': 'Likewise, we apply the chain rule to the remaining terms and get These! Bam! Now we can simplify the equation.', 'start': 856.968, 'duration': 15.13}, {'end': 881.346, 'text': 'These zeros go away, and we can pull the sigma squared out and add the numerators together.', 'start': 873.7, 'duration': 7.646}, {'end': 884.748, 'text': 'Then we can combine the measurements.', 'start': 883.087, 'duration': 1.661}, {'end': 889.032, 'text': "And lastly, we can combine the mu's.", 'start': 886.61, 'duration': 2.422}, {'end': 895.77, 'text': 'Thus, this is the derivative of the log-likelihood function with respect to mu.', 'start': 890.728, 'duration': 5.042}, {'end': 904.675, 'text': "Double bam! Now let's take the derivative of the log-likelihood function with respect to sigma.", 'start': 897.491, 'duration': 7.184}, {'end': 909.737, 'text': 'This derivative is the slope function for the log of the likelihood curve.', 'start': 905.875, 'duration': 3.862}, {'end': 914.88, 'text': "And we'll use it to find the peak, which is hard to see in this graph.", 'start': 911.378, 'duration': 3.502}], 'summary': 'Derivative calculations applied to log-likelihood function with respect to mu and sigma, simplifying equations and finding slope function for likelihood curve.', 'duration': 66.754, 'max_score': 848.126, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/Dn6b9fCIUpM/pics/Dn6b9fCIUpM848126.jpg'}, {'end': 1030.815, 'src': 'embed', 'start': 998.683, 'weight': 2, 'content': [{'end': 1004.907, 'text': 'We can simplify the equation by omitting the zero, leaving the second term the same,', 'start': 998.683, 'duration': 6.224}, {'end': 1011.676, 'text': 'and we can pull the sigma cubed out and add up the remaining terms.', 'start': 1006.451, 'duration': 5.225}, {'end': 1018.823, 'text': 'Double bam! At long last, here are the two derivatives.', 'start': 1013.298, 'duration': 5.525}, {'end': 1028.332, 'text': 'To find the maximum likelihood estimate for mu, we need to solve for where the derivative with respect to mu equals zero,', 'start': 1020.224, 'duration': 8.108}, {'end': 1030.815, 'text': 'because the slope is zero at the peak of the curve.', 'start': 1028.332, 'duration': 2.483}], 'summary': 'Simplify equation, find derivatives, solve for maximum likelihood estimate.', 'duration': 32.132, 'max_score': 998.683, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/Dn6b9fCIUpM/pics/Dn6b9fCIUpM998683.jpg'}, {'end': 1133.771, 'src': 'embed', 'start': 1103.684, 'weight': 1, 'content': [{'end': 1106.146, 'text': 'Multiply both sides by sigma squared.', 'start': 1103.684, 'duration': 2.462}, {'end': 1111.411, 'text': 'Divide both sides by n.', 'start': 1107.727, 'duration': 3.684}, {'end': 1113.633, 'text': 'And take the square root of both sides.', 'start': 1111.411, 'duration': 2.222}, {'end': 1123.204, 'text': 'And, at long last, we see that the maximum likelihood estimate for sigma is the standard deviation of the measurements.', 'start': 1115.519, 'duration': 7.685}, {'end': 1133.771, 'text': 'Thus, we use the formula for the standard deviation to determine the width of the normal curve that, given the data, has the maximum likelihood.', 'start': 1124.985, 'duration': 8.786}], 'summary': 'The maximum likelihood estimate for sigma is the standard deviation of the measurements.', 'duration': 30.087, 'max_score': 1103.684, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/Dn6b9fCIUpM/pics/Dn6b9fCIUpM1103684.jpg'}], 'start': 743.906, 'title': 'Derivatives in maximum likelihood estimation', 'summary': 'Explains the process of taking the derivative of the log likelihood function to find the peak, involving the chain rule and simplification. it also discusses the maximum likelihood estimates for mu and sigma, which are the mean and standard deviation of the measurements, respectively.', 'chapters': [{'end': 889.032, 'start': 743.906, 'title': 'Derivative of log likelihood function', 'summary': 'Explains the simplification of the log likelihood function and the process of taking its derivative with respect to mu in order to find the peak, where the slope equals zero, involving the application of the chain rule and simplification of the equation.', 'duration': 145.126, 'highlights': ['The derivative with respect to mu is used to find the peak of the log likelihood curve, where the slope equals zero.', 'The application of the chain rule is demonstrated in taking the derivative of the term containing mu, resulting in the simplification of the equation.', "The process involves combining terms, canceling zeros, pulling out sigma squared, adding numerators, and combining measurements and mu's."]}, {'end': 1182.593, 'start': 890.728, 'title': 'Maximum likelihood estimation', 'summary': 'Discusses the derivatives of the log-likelihood function with respect to mu and sigma, leading to the maximum likelihood estimates for mu and sigma, which are the mean and standard deviation of the measurements, respectively.', 'duration': 291.865, 'highlights': ['The maximum likelihood estimate for sigma is the standard deviation of the measurements, demonstrated by setting the derivative of the log-likelihood function with respect to sigma to zero, simplifying the equation, and deriving the sigma estimate.', 'The maximum likelihood estimate for mu is the mean of the measurements, derived by setting the derivative of the log-likelihood function with respect to mu to zero and solving for mu.', 'The chapter outlines the process of finding the peak of the likelihood curve by taking the derivatives of the log-likelihood function with respect to mu and sigma, simplifying the equations, and demonstrating that the peak occurs at the center of the normal distribution and the standard deviation of the data.']}], 'duration': 438.687, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/Dn6b9fCIUpM/pics/Dn6b9fCIUpM743906.jpg', 'highlights': ["The process involves combining terms, canceling zeros, pulling out sigma squared, adding numerators, and combining measurements and mu's.", 'The maximum likelihood estimate for sigma is the standard deviation of the measurements, demonstrated by setting the derivative of the log-likelihood function with respect to sigma to zero, simplifying the equation, and deriving the sigma estimate.', 'The maximum likelihood estimate for mu is the mean of the measurements, derived by setting the derivative of the log-likelihood function with respect to mu to zero and solving for mu.', 'The derivative with respect to mu is used to find the peak of the log likelihood curve, where the slope equals zero.', 'The application of the chain rule is demonstrated in taking the derivative of the term containing mu, resulting in the simplification of the equation.', 'The chapter outlines the process of finding the peak of the likelihood curve by taking the derivatives of the log-likelihood function with respect to mu and sigma, simplifying the equations, and demonstrating that the peak occurs at the center of the normal distribution and the standard deviation of the data.']}], 'highlights': ['The likelihood of the new curve with mu equals 30 and sigma equals 2, given the data, is 0.12, demonstrating the process of calculating likelihood with specific values. (Quantifiable data: likelihood value of 0.12)', 'The likelihood of the curve with mu equals 30 and sigma equals 2, given the data, is 0.03, indicating the initial likelihood calculation process with specific values. (Quantifiable data: likelihood value of 0.03)', 'The mouse weighs 32 grams, serving as the specific data used for the likelihood calculation example. (Quantifiable data: weight of the mouse is 32 grams)', 'The likelihood function equals the product of the likelihood functions for the n individual measurements, with n data points, and we multiply together all n individual likelihood functions.', 'Deriving the maximum likelihood estimate for mu by finding where the derivative equals zero with respect to mu, treating sigma as a constant, and doing the same for sigma by finding where the derivative equals zero with respect to sigma, treating mu as a constant.', 'The maximum likelihood estimate for sigma is the standard deviation of the measurements, demonstrated by setting the derivative of the log-likelihood function with respect to sigma to zero, simplifying the equation, and deriving the sigma estimate.']}