# Clarification on random variables?

1. If $X$ and $Y$ are dependent random variables, then it is possible that $Var(X+Y) > Var(X) + Var(Y)$.

I only know that the two are equal for independent random variables; for dependent variables, would this be the case?

1. According to forecasts, the end-of-year value in dollars of IBM stock has variance 10. If an investor holds a portfolio containing 5 shares of IBM stock and 240 dollars of idle cash, what is the variance in the end-of-year value in dollars of his portfolio?

I would assume it's 50, as variance is additive?

1. If X and Y are random variables such that $P(X=0)=0.5$ and $P(Y=0)=0.1$, then is $P((X+Y)/2=0)$ equal to $P(X=0)/2 + P(Y=0)/2 = 0.3$?

I don't believe that it's possible to add probabilities like this, would I multiply instead?

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In general, $\operatorname{var}(X+Y) = \operatorname{var} X + \operatorname{var} Y + \operatorname{Cov}(X,Y)$, where the last term is the covariance. If $X$ and $Y$ are positively correlated ("$X$ and $Y$ increase or decrease together"), then this term will be positive. – Clement C. Mar 5 at 23:42
@Clement You are missing a factor of $2$ in front of $cov(X,Y)$ or $cov(X,Y)+(cov(X,Y))^T$ in the multi-dimensional case. – A.S. Mar 5 at 23:59
Oh, true -- thanks for pointing out. And I was going to write something along the line of "it's easy to remember, as it mimics the formula for $(a+b)^2=a^2+b^2+2ab$ with the cross-term"... yes, easy to remember indeed. – Clement C. Mar 6 at 0:05

1: $\text{Var}(Z) = \text{E}\Big[\big(Z - \text{E}(Z)\big)^2\Big]$. It follows that \begin{align} \text{Var}(X+Y) &= \text{E}\bigg[\big(X+Y - \text{E}(X) - \text{E}(Y)\big)^2\bigg] \\ &= \text{E}\bigg[\big(X - \text{E}(X)\big)^2 + \big(Y - \text{E}(Y)\big)^2 + 2\big(X - \text{E}(Y)\big)\big(Y-\text{E}(Y)\big)\bigg] \\ &= \text{Var}(X) + \text{Var}(Y) + 2\text{Cov}(X,Y). \end{align} So if the covariance is positive, the variance of $X+Y$ exceeds $\text{Var}(X) + \text{Var}(Y)$.

2: No, it's 250. You have that $\text{Var}(aX) = a^2\text{Var}(X)$, and therefore the variance of $5$ stocks with variance $10$ is $5^2\cdot 10 = 250$. Technically, you then need to consider the variance of those 240 dollars. But since their value doesn't have any uncertainty, that variance is zero.

3: You're correct that the answer is "no". Its not even possible to compute the probability precisely, because $X+Y$ can be zero even if $X$ and $Y$ are both non-zero. So $P(X=0)\cdot P(Y=0)$ is only a lower bound for the probability $P(X+Y=0)$.

Edited: The answer previously stated that 50 is the correct answer for question 2, which was wrong.

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@Teepeemm Uh, yes, of course. Fixed. – fgp Mar 6 at 8:40

Suppose that $Y=X$. Then the variance of $X+Y$ is the variance of $2X$, which is $4$ times the variance of $X$.

So for example if $X=1$ if when tossing a fair coin we get a head, and $X=0$ otherwise, and $Y=X$, then $\text{Var}(X+Y)\gt \text{Var}(X)+\text{Var}(Y)$.

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