Clarification on random variables? 
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*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?


*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?


*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?
 A: 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 $\text{Var}(X+Y)\gt \text{Var}(X)+\text{Var}(Y)$.
A: 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. 
