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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.

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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.) – whuber♦ Feb 18 '12 at 22:02
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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: $\mathbb{V}(\bar{X}) = \frac{1}{n^2}\sum^n_{i=1}{\mathbb{V}(X_i)}$ – John Smith Feb 18 '12 at 22:17
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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 $\frac{1}{n^2}$.) – whuber♦ Feb 18 '12 at 22:21
ok thanks. this was very helpful. – John Smith Feb 18 '12 at 22:21
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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 $\mathbb{V}(\bar{X}) = \frac{1}{n^2}\sum^n_{i=1}{\mathbb{V}(X_i)} = \frac{1}{n^2}n\frac{0.1*0.5}{(0.1+0.5)^2(0.1+0.5+1)} = \frac{1}{n}\frac{0.05}{(0.36)(1.6)}$ – John Smith Feb 18 '12 at 22:42
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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.

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Yea, I saw that and thought the covariance term was 0 because the random variables were drawn iid. – John Smith Feb 19 '12 at 1:36
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The formula for the variance of $X+Y$ is wrong. It should be $$\mathbb V(X+Y) = \mathbb V(X) + \mathbb V(Y) + 2\text{COV}(X, Y).$$ – Dilip Sarwate Feb 19 '12 at 1:52
John you're correct, that's exactly why it is 0. Dilip thanks for catching the typo, will fix it now. – sparc_spread Feb 19 '12 at 2:44

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