Expected value of sum of two discrete random variables

Question

Correct Answer: 1.1

My work: I'm not sure, but I tried to find the value for $c$ by summing joint pmf over all possible values of x and y, in order to use Law of Unconscious probabilist to calculate E[Y-X]. I am stuck on calculating $c$ however...

Answer 1

You will first have to compute $c$ using that the probabilities of all your outcomes should sum to 1, i.e. $$ \sum_{y=0}^3 \sum_{x=0}^y c(x + 2y) = 1,$$ which means that $$c = \left[\sum_{y=0}^3 \sum_{x=0}^y (x + 2y)\right]^{-1}$$ which in our case can be computed by WolframAlpha to be $\frac{1}{50}$.

Answer 2

The quantity that we want is $$\operatorname{E}[Y-X],$$ the expected value of the total number of tornadoes minus the number of tornadoes that result in at least 50 million in losses. By linearity of expectation, we know that $$\operatorname{E}[Y-X] = \operatorname{E}[Y] - \operatorname{E}[X].$$ By definition, we have $$\operatorname{E}[Y] = \sum_{y=0}^3 y \Pr[Y = y],$$ and similarly for $X$. But since we are only given the joint mass function, i.e. $$p_{X,Y}(x,y) = \Pr[(X = x) \cap (Y = y)],$$ in order to compute the marginal probabilities, we need to write $$\Pr[Y = y] = \sum_{x=0}^3 \Pr[(X = x) \cap (Y = y)].$$ So for example, $$\Pr[Y = 2] = \Pr[(X = 0) \cap (Y = 2)] + \Pr[(X = 1) \cap (Y = 2)] + \Pr[(X = 2) \cap (Y = 2)],$$ noting that $\Pr[(X = 3) \cap (Y = 2)] = 0$.

So, our entire problem-solving strategy may be better executed by using a table; e.g. $$\begin{array}{c|c|c|c|c|c} & X = 0 & X = 1 & X = 2 & X = 3 & \text{Total} \\ \hline Y = 0 & 0 & 0 & 0 & 0 & 0 \\ \hline Y = 1 & 2c & 3c & 0 & 0 & 5c \\ \hline Y = 2 & 4c & 5c & 6c & 0 & 15c \\ \hline Y = 3 & 6c & 7c & 8c & 9c & 30c \\ \hline \text{Total} & 12c & 15c & 14c & 9c & 50c \end{array}$$ This tells us that, in order for the total probability to sum up to $1$, we must have $50c = 1$. This also then gives us the marginal probabilities for $X$ and for $Y$, as shown in the row total and column total, from which we easily compute the required expected values.