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)
Please could you explain at 3:30 why Var(xi) = σ^2 rather than s^2 ? Isn't σ^2 the population variance, but xi is defined from i=1 to n so isn't it a sample? Or is the population only defined from i=1 to n?
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.
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.
Yes. As discussed in this video, var(xbar) = sigma^2/n. Now, you also have a constant, 1/2, since you want to find var(1/2 * xbar). In general, if "a" is a constant, and "x" is random, then var(ax) = a^2 * var (x). Therefore var(1/2 * xbar ) = 1/4 * var(xbar) = 1/4 * sigma^2/n
I think the proof is wrong since proposition Var(Xi) = sigma^2 i={1,2,...n} is not correct, keeping in mind that sigma^2 is actually the population variance, which by definition differs from the deviation of every single Xi (except if Xi=c for all i)
On the other hand I think that your confusion comes from assuming that mentioned above proposition is the same as THE MEAN OF Var(xi) for all i={1,2,...,n} which ofc is equal to squared sigma.
@@nicolas0851 no, the video is correct. This calculation of VAR(Xbar) assumes that X_i , i = 1,2,3 are i.i.d and by definition of i.i.d they have the same variance sigma^2
In this video, there is only 1 random variable, X, therefore, we do not need an independence assumption. x1, x2, ... xn are different observations for the random variable X. This website might be helpful in explaining random variables in more detail: www.investopedia.com/terms/r/random-variable.asp#:~:text=Key%20Takeaways,value%20in%20a%20continuous%20range).
Are you asking for the proof for Chebyshev's Inequality? Can you provide more details for this question? I am not sure why we would use Chebyshev's Inequality to find the variance of a sample mean? Though, I feel like I am missing something in this question.... usually we use Chebyshev's Inequality to tell us something about the probability of randomly sampling a value within a specified number of standard deviations from the mean... For example, Chebyshev's inequality tells us that at least 75% of values (regardless of the distribution for those values) are within 1 standard deviation from the mean.
@@Stats4Everyone Hi! Sorry it took me so long to reply. It turned out the question was faulty, what the Professor actually wanted was an upper limit approximation for the variance. Thanks anyway!
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Thank you for improving my poor maths, i have been struggling to understand this in "Tests of significance"in biostatistics
Had this is my assignment and didn't know how to do it, but now I get it, Thanks a bunch.
Thank you a lott
I was struggling to understand this
Lovely explanation, very clear. Only a doubts: which is the easiest demostration that Var(ay)=a^2var(y)? (I just subscribed, thank You)
Thank you for this comment! This is an excellent question! Here is a video: ruclips.net/video/5ywOKf25VJ0/видео.html
@@Stats4Everyone thank You to answer to Me 😘 . Ciao from Italy
Thank you very much 😊
This has troubled me for so long!
Glad you found this video to be helpful!
Please explain why the constant gets squared Var(ay)=a^2Var(y)
Great question! Here is a video: ruclips.net/video/5ywOKf25VJ0/видео.html
Thank you for the video, very helpful. What if the equation is like Var (X ̅) = 1/n ∑Var (Xi) ?
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)
Please could you explain at 3:30 why Var(xi) = σ^2 rather than s^2 ? Isn't σ^2 the population variance, but xi is defined from i=1 to n so isn't it a sample?
Or is the population only defined from i=1 to n?
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.
Am I right, that this works only when the distribution of the variables are the same?
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.
Hi
Thankyou for this insightful video! Can you please share which book or research paper you referred to?
hi sis, im sorry, I want to ask, what if the question asked is Var(Xbar/2)? is the output a sigma^2/4n?
are you willing to explain to me?
Yes. As discussed in this video, var(xbar) = sigma^2/n. Now, you also have a constant, 1/2, since you want to find var(1/2 * xbar). In general, if "a" is a constant, and "x" is random, then var(ax) = a^2 * var (x). Therefore var(1/2 * xbar ) = 1/4 * var(xbar) = 1/4 * sigma^2/n
I think the proof is wrong since proposition Var(Xi) = sigma^2 i={1,2,...n} is not correct, keeping in mind that sigma^2 is actually the population variance, which by definition differs from the deviation of every single Xi (except if Xi=c for all i)
On the other hand I think that your confusion comes from assuming that mentioned above proposition is the same as THE MEAN OF Var(xi) for all i={1,2,...,n} which ofc is equal to squared sigma.
Pls, correct me if I'm wrong, I'd really appreaciate it.
Last but not least, I know that surely I'm wrong due to some misconception of mine haha.
@@nicolas0851 no, the video is correct. This calculation of VAR(Xbar) assumes that X_i , i = 1,2,3 are i.i.d and by definition of i.i.d they have the same variance sigma^2
Mam, Do we need to assume that the random vars are independent?
In this video, there is only 1 random variable, X, therefore, we do not need an independence assumption. x1, x2, ... xn are different observations for the random variable X. This website might be helpful in explaining random variables in more detail: www.investopedia.com/terms/r/random-variable.asp#:~:text=Key%20Takeaways,value%20in%20a%20continuous%20range).
Great video! Do you know how one could get that formula by starting with Chebyshevs-inequation?
Are you asking for the proof for Chebyshev's Inequality? Can you provide more details for this question? I am not sure why we would use Chebyshev's Inequality to find the variance of a sample mean? Though, I feel like I am missing something in this question.... usually we use Chebyshev's Inequality to tell us something about the probability of randomly sampling a value within a specified number of standard deviations from the mean... For example, Chebyshev's inequality tells us that at least 75% of values (regardless of the distribution for those values) are within 1 standard deviation from the mean.
@@Stats4Everyone Hi! Sorry it took me so long to reply. It turned out the question was faulty, what the Professor actually wanted was an upper limit approximation for the variance. Thanks anyway!