If the joint pmf of $X,Y$ is $f(x,y) = \frac{1}{58} (x^2 + y^2),$ find the expectation and variance of $X$.

Question

Random variables $X$ and $Y$ have the joint probability distribution: $$f(x,y) = \frac{1}{58} (x^2 + y^2),$$ when $x= 1,2,3$ and $y= 1,3$ and $f(x,y)=0$, otherwise.
a) Find $E(X)$ and $E(Y).$
b) Find $\text{Var}(X)$ and $\text{Var}(Y).$

For the expectations, do I take the integral with respect to just $x$ for $E(X)$ and multiply it by $x$? Or is that wrong? I don't have any clue how to find the variances. Any ideas?

Answer 1

You need to use double sums. (The supports for $X, Y$ are discrete values). EG: $$\begin{align}\mathsf E(X) ~=~& \sum\limits_{x\in\{1,2,3\}}\sum\limits_{y\in\{1,3\}} x\cdot\dfrac{x^2+y^2}{58} \\[1ex] =~& \dfrac 1{58} \sum\limits_{x\in\{1,2,3\}}(2x^3+x\sum\limits_{y\in\{1,3\}}y^2) \\[1ex]=~& \dfrac 1{58}(2(1+8+27)+(1+2+3)(1+9)) \\[1ex]=~& \dfrac {66}{29} \end{align}$$

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Reality Check: $$\begin{align}\mathsf E(1) ~=~& \sum\limits_{x\in\{1,2,3\}}\sum\limits_{y\in\{1,3\}} 1\cdot\dfrac{x^2+y^2}{58} \\[1ex] =~& \dfrac 1{58} (2\sum\limits_{x\in\{1,2,3\}}x^2~+~3\sum\limits_{y\in\{1,3\}}y^2) \\[1ex]=~& \dfrac 1{58}(2(1+4+9)+3(1+9)) \\[1ex]=~& 1 \end{align}$$

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Variances are found by $\mathsf {Var}(X) = \mathsf {E}(X^2) - \mathsf E(X)^2\\\mathsf {Var}(Y) = \mathsf {E}(Y^2) - \mathsf E(Y)^2$.

And of course, $$\begin{align}\mathsf E(X^2) ~=~& \sum\limits_{x\in\{1,2,3\}}\sum\limits_{y\in\{1,3\}} x^2\cdot\dfrac{x^2+y^2}{58}\end{align}$$

and so on

Answer 2

Don't let the $f(x,y)$ fool you. You have a discrete distribution, since the possible values of $X$ are $\{1,2,3\}$ and the possible values of $Y$ are $\{1,3\}$.

Essentially, you're being asked to find the marginal densities of $X$ and $Y$.

One way to clearly see this is to compute the joint distribution table. It is not necessary, but it helps to see everything; let the rows be the values of $X$, and the columns be the values of $Y$,

\begin{array}{r|c|c|l} f_{X,Y}(x,y)&1&3&P(X=x)\\\hline 1&\dfrac{2}{58}&\dfrac{10}{58}&\dfrac{12}{58}\\\hline 2&\dfrac{5}{58}&\dfrac{13}{58}&\dfrac{18}{58}\\\hline 3&\dfrac{10}{58}&\dfrac{18}{58}&\dfrac{28}{58}\\\hline P(Y=y)&\dfrac{17}{58}&\dfrac{41}{58}&1 \end{array} Now, that we have the marginal distributions of $X$ and $Y$, you can compute the expectations and variances as usual. For example, $$E[X] = 1\cdot\frac{12}{58}+2\cdot\frac{18}{58}+3\cdot\frac{28}{58} = \frac{66}{29}.$$