5
$\begingroup$

The question is

Let $X_1,...,X_n$ be drawn iid from $Beta(0.1,0.5)$. Let $\bar{X} = \frac{1}{n}\sum^n_{i=1} X_i$.

a) Derive $\mathbb{E}(\bar{X})$ and $\mathbb{V}(\bar{X})$

I know how to get the expectation and variance of a random variable from a given distribution. But I'm not how to get the expectation and variance of $\bar{X}%$. Not sure where to being actually.

$\endgroup$
5
  • 4
    $\begingroup$ What rules do you know that might enable you to compute the expectation and variance of a sum of random variables or a constant multiple of a random variable? (You can look up the expectation and variance of a Beta distribution: Wikipedia lists them, for example.) $\endgroup$
    – whuber
    Feb 18, 2012 at 22:02
  • 2
    $\begingroup$ So I can change $\mathbb{E}(\bar{X})$ to $\sum^n_{i=1}\mathbb{E}(\bar{X_i})$, which I can calculate since I know the expected value of a single random variable from the Beta distribution. Similarly for the variance, I can do: $\mathbb{V}(\bar{X}) = \frac{1}{n^2}\sum^n_{i=1}{\mathbb{V}(X_i)}$ $\endgroup$
    – John Smith
    Feb 18, 2012 at 22:17
  • 2
    $\begingroup$ Yes, that is correct: it called the "linearity of expectation." Variances enjoy a comparable relationship with linear combinations, although a little more complicated, but the same idea will serve you in good stead. (I see you have just edited your comment to include the variance rule; I'm glad you correctly converted the $\frac{1}{n}$ to $\frac{1}{n^2}$.) $\endgroup$
    – whuber
    Feb 18, 2012 at 22:21
  • $\begingroup$ When you obtain an answer you're confident is correct, please post it as a reply here so we have it for the record. :-) $\endgroup$
    – whuber
    Feb 18, 2012 at 22:23
  • 1
    $\begingroup$ So I got $\mathbb{E}(\bar{X}) = \frac{1}{n}\sum^n_{i=1}{\mathbb{E}(X_i)} = \frac{1}{n}n\frac{0.1}{0.1+0.5} = \frac{1}{6} $ and $\mathbb{V}(\bar{X}) = \frac{1}{n^2}\sum^n_{i=1}{\mathbb{V}(X_i)} = \frac{1}{n^2}n\frac{0.1*0.5}{(0.1+0.5)^2(0.1+0.5+1)} = \frac{1}{n}\frac{0.05}{(0.36)(1.6)}$ $\endgroup$
    – John Smith
    Feb 18, 2012 at 22:42

1 Answer 1

5
$\begingroup$

I got the same answer as you. However, be aware of the following:

$ \mathbb{V}(X + Y) = \mathbb{V}(X) + \mathbb{V}(Y) + 2\text{COV}(X,Y) $

For a long summation of variances, like the one you did, it becomes:

$ \mathbb{V}(\sum^n_{i=1}X_i)=\sum^n_{i=1}\sum^n_{j=1}\text{COV}(X_i,X_j) $

This is the "more complicated" part that whuber mentioned :-). Make sure you understand what happens to the covariances and why. Let me know if you need help there.

$\endgroup$
3
  • 1
    $\begingroup$ Yea, I saw that and thought the covariance term was 0 because the random variables were drawn iid. $\endgroup$
    – John Smith
    Feb 19, 2012 at 1:36
  • 1
    $\begingroup$ The formula for the variance of $X+Y$ is wrong. It should be $$\mathbb V(X+Y) = \mathbb V(X) + \mathbb V(Y) + 2\text{COV}(X, Y).$$ $\endgroup$ Feb 19, 2012 at 1:52
  • $\begingroup$ John you're correct, that's exactly why it is 0. Dilip thanks for catching the typo, will fix it now. $\endgroup$ Feb 19, 2012 at 2:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.