Відео

Because generally the problem has you assume that each birthday is equally likely in order to find the probability, but here I am simply supposed to prove that the probability is maximized under that assumption

Good question. Thank you for posting this comment. ybar is a constant (it is the same number for all i), therefore, we can factor it as follows: sum (xi-xbar)*ybar = ybar * sum(xi-xbar) sum(xi - xbar) = 0 . For proof, please see this video: ua-cam.com/video/oyjkFmNMMKA/v-deo.html Therefore, sum (xi-xbar)*ybar = ybar * sum(xi-xbar) = ybar * 0 In contrast, yi can not be factored, because it is not a constant, and changes with i. Therefore, sum (xi-xbar)*yi is not zero, because for each i, xi-xbar is being multiplied by yi.

In your course, if the estimator for sample variance? For example, if you are estimating a mean, the unbiased estimator would have n in the denomiator... Though for sample variance, the proof that I provide in this video is correct. Here is another source that might be helpful: en.wikipedia.org/wiki/Bias_of_an_estimator#:~:text=Sample%20variance,-Main%20article%3A%20Sample&text=Dividing%20instead%20by%20n%20%E2%88%92%201,results%20in%20a%20biased%20estimator.

Thank you for the comment and encouragement :-)

At min 3:15, I have written a similar equation, though your equation is missing that 1/n is squared (maybe that was a typo?). Here is the correct equation: Var ( Xbar ) = (1/n)^2 ∑ Var(Xi)

Good question. Thanks for this post. Towards the beginning of this video, I discuss this a little, though I can try to elaborate here. I will break down your questions below: 1. why Var(xi) = σ^2 rather than s^2 ? s^2 is an estimator for the population variance σ^2. In other words, s^2 estimates σ^2 using the sample x1, x2,...xn. Just like x-bar is an estimator for the population mean, mu. By definition, X is a random variable with mean mu, and variance σ^2. While we may not know the variance, σ^2, and mean, mu, all xi's sampled from the random variable X will have variance, σ^2, and mean, mu. Using the xi's sampled from the random variable X, we can estimate the variance, σ^2, using s^2, and the mean, mu, using the sample mean, x-bar. 2. Isn't σ^2 the population variance, but xi is defined from i=1 to n so isn't it a sample You are correct that σ^2 the population variance. Also, you are correct that xi, for i=1,2..n is a sample. xi is a sample observation of the random variable x, which has variance σ^2. 3. Or is the population only defined from i=1 to n? The population is not only defined only for i=1 to n. It does not matter how many times we sample from the random variable x, the variance will always be σ^2 by definition of x. I hope this helps. If you have any follow-up questions, please let me know.

@@Stats4Everyone Thank you for the reply. So each random variable Xi has variance σ^2 before it is defined, I think I understand now.

Glad it was helpful!

Great to hear! I'm glad it was helpful!

Yes. Good question. Most importantly, all xi, for i=1..n, must have the same variance, σ^2, which is discussed a little towards the beginning of the video, near min 0.20.

Good question. Thanks for this comment. The population mean is not known. Therefore, we can not say for a fact that the "sample mean is higher than population mean". Since the sample mean is a random number (based on random sampling), the sample mean is occasionally more than the population mean, and occasionally less than the population mean, though on average, if we were to repeatedly take samples, the sample mean will be equal to the population mean. This is the law of large numbers (en.wikipedia.org/wiki/Law_of_large_numbers) In this video example, we want to know if a college student spends more than 20 hours per week studying. Therefore, we want to know if the population mean is more than 20. This is our question: Is mu > 20? To answer this question, we obtained a random sample, and found that the sample mean is 23 hours per week. The sample is random. A different sample of 36 students would yield a different sample mean. I hope this makes sense and is helpful!

Glad to hear! Thanks for the comment!

Great! Glad that this video was helpful for your classwork :-)

Thanks for this comment! I have broken down a response below: 1. Is there difference of E[yi] and y_bar Yes there is a difference. ybar is an estimator for the expected value of y, E(y). ybar is a sample average, whereas the expected value of y, E(y) is the population average. Also, ybar is not conditional on the value of x, rather it is an average over all the different values of x. In a hypothetical example, if we were using tree height, x, to model tree age, y, then ybar is the sample average tree age of all the sampled trees, regardless of their height. E(y) is the population average tree age of all trees in the population, regardless of their height. E(yi) is the expected value of yi. In our hypothetical example, E(yi) is the expected tree age for a particular tree height. In other words, E(yi) is the average tree age for a particular tree height (“expected” and “average” are synonymous). yi is a single observed tree age at a particular tree height. We could observe and sample several values trees at the exact same height, and then average their ages to estimate E(yi). 2. similar like E[epsilon_i] and epsilon_bar Not really… We assume that E[epsilon_i] is zero as a model assumption. This means that the model is explaining the variability in y, and anything else making y vary is just random noise. In the hypothetical example, this means that tree height is the only predictor for tree age and everything else is either controlled or not important in explaining why all trees are not the exact same age (note that this assumption is definitely a stretch and unrealistic in this example…and yet people might run this regression without much thought about whether the model assumptions are reasonable…which is not good). Epsilon_bar. It is important to define this idea. Usually, people use lower case e to denote the sample model error term, since epsilon_i is not observed (unlike y_i, which is observed). The sample average of the model error term, e, is zero by definition. It is a fun exercise to prove the following: If, ei = yihat - yi where yihat = bo + b1xi then, sum ei = 0 therefore e-bar = 0 To do this proof, plug in the estimators bo and b1. I hope this helps! Thanks again for the thoughtful comment!

So happy to hear that this video was helpful!

So happy to hear that this video was helpful! :-)

Thanks so much for this feedback and encouragement!

Thanks for this comment! Here is a video that might be helpful: ua-cam.com/video/5Ezg9UXIyZs/v-deo.htmlsi=IkU9c9jmBlIorGw6 First, let's think about Yi, which is a random number. To be a random number, we can think of that as a number that when sampled is randomly not the same number. For example, suppose I am interested in knowing the IQ of people in America. Not everyone's IQ is the same. IQ varries randomly from person to person. Actually, we could measure the IQ of the same person twice and likely get two different numbers. Generally, things that are "measured" are random numbers because exact measurements are very difficult/impossible. A non random number is constant. Age might be a good example of a nonrandom X value. If we know someone's birthday, Age can be calculated exactly.

@@Stats4Everyone Thank you so much for your time and detailed reply! Yup very helpful! So the independent sample data is nonrandom per samples that collected. But the dependent data (the one that measured, the one that we study the relation to the sample's data) are random, cuz even tho we draw the same sample, but the dependent variable we study still subject to vary within a certain distribution. Really appreciate it! Wish all the best to you and your channel!