# Find the mean and variance of $Y=\sum\limits_{i=1}^3 X_i$

Let $X_1$, $X_2$, $X_3$ be random variable having same density function: $f(x)=2x/9$ when $0 < x < 3$ and $f(x)=0$ elsewhere.
Let $Y=X_1+X_2+X_3$.
I need to find the mean and variance of $Y$.
I know how to calculate mean, but for the variance do I need covariance of $X_1$, $X_2$ and $X_3$?
How can I solve for them?

• Does you assignment state that $X_1$, $X_2$ and $X_3$ are independent? – Sasha May 2 '12 at 0:49
• No, that's the problem. There is no more information about random variables. – bernio May 2 '12 at 0:54
• Then there's not enough information to solve the problem. You might be able to find bounds for the variance, though. – Robert Israel May 2 '12 at 8:12
• The minimum possible variance is $0$. This would be the case, for example, if $X_1 = \sqrt{2/3} (Y_1 - Y_2/2 - Y_3/2)$, $X_2 = \sqrt{2/3} (-Y_1/2 +Y_2 - Y_3/2)$, $X_3 = \sqrt{2/3} (-Y_1/2 - Y_2/2 + Y_3)$ where $Y_1$, $Y_2$, $Y_3$ are independent random variables with the given distribution. The maximum possible variance is $9$ times the variance of $X_1$, obtained when $X_1, X_2, X_3$ are all the same random variable. – Robert Israel May 2 '12 at 8:48

By the Linearity of Expectation, $$\mathrm{E}(X_1+X_2+X_3)=\mathrm{E}(X_1)+\mathrm{E}(X_2)+\mathrm{E}(X_3)$$ However, using the Linearity of Expectation and $\mathrm{Var}(X)=\mathrm{E}(X^2)-\mathrm{E}(X)^2$, we get \small\begin{align} &\mathrm{Var}(X_1+X_2+X_3)\\ &=\mathrm{E}\left((X_1+X_2+X_3)^2\right)-\mathrm{E}(X_1+X_2+X_3)^2\\ &=\mathrm{E}(X_1^2)-\mathrm{E}(X_1)^2+\mathrm{E}(X_2^2)-\mathrm{E}(X_2)^2+\mathrm{E}(X_3^2)-\mathrm{E}(X_3)^2\\ &+2\mathrm{E}(X_1X_2)-2\mathrm{E}(X_1)\mathrm{E}(X_2)+2\mathrm{E}(X_2X_3)-2\mathrm{E}(X_2)\mathrm{E}(X_3)+2\mathrm{E}(X_3X_1)-2\mathrm{E}(X_3)\mathrm{E}(X_1)\\ &=\mathrm{Var}(X_1)+\mathrm{Var}(X_2)+\mathrm{Var}(X_3)+2\mathrm{Cov}(X_1,X_2)+2\mathrm{Cov}(X_2,X_3)+2\mathrm{Cov}(X_3,X_1) \end{align} where $$\mathrm{Cov}(X,Y)=\mathrm{E}(XY)-\mathrm{E}(X)\mathrm{E}(Y)$$ Therefore, using what you have, you can compute the mean of $Y$, but to compute the variance of $Y$, you need to know the covariances of $X_1$, $X_2$, and $X_3$. On the other hand, if $X_1$, $X_2$, and $X_3$ are independent, then their covariances are $0$, and then you can compute $\mathrm{Var}(Y)$: $$\mathrm{Var}(X_1+X_2+X_3)=\mathrm{Var}(X_1)+\mathrm{Var}(X_2)+\mathrm{Var}(X_3)$$

Just to expand on Robert Israel's answer - we have a formula for the variance of a random variable which is a sum of random variables:

$$\mathrm{Var}{(X_{1}+X_{2}+...+X_{n})}=\mathrm{Var}X_{1}+\mathrm{Var}X_{2}+...+\mathrm{Var}X_{n}+2\sum_{i<j}\mathrm{Cov}(X_{i},X_{j})$$

$$\mathrm{Var}Y=\mathrm{Var}(X_{1}+X_{2}+X_{3})=\mathrm{Var}X_{1}+\mathrm{Var}X_{2}+\mathrm{Var}X_{3}+2\left[\mathrm{Cov}(X_{1},X_{2})+\mathrm{Cov}(X_{1},X_{3})+\mathrm{Cov}(X_{2},X_{3})\right]$$

And that is why, as Robert Israel said, we need to have information about the covariance of these three random variables. IF they would be independent, we would have:

$$\mathrm{Cov}(X_{1},X_{2})=\mathrm{Cov}(X_{1},X_{3})=\mathrm{Cov}(X_{2},X_{3})=0$$

and hence the variance of $Y$ would be reduced to

$$\mathrm{Var}Y=\mathrm{Var}X_{1}+\mathrm{Var}X_{2}+\mathrm{Var}X_{3}$$

but then again, that has to be explicitly stated.

• Heh, two very similar answers, just 17 seconds apart. It is almost as impressive as the three almost identical answers to this question within 13 seconds. (+1) – robjohn Oct 20 '12 at 8:15
• @robjohn - Yep, that's pretty impressive. Although your answer actually explains why the formula that both of us gave is true - which I guess, in the end, should be beneficial for the OP... :) – Johnny Westerling Oct 20 '12 at 8:20

Yes, you need the covariances, and without them (or some other information about the problem) you can't go any farther. For example, perhaps $X_1$, $X_2$ and $X_3$ are independent. But if so that should be stated as part of the problem.