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Your evil probability professor has an urn with 9 balls: 2 red, 3 white and 4 blue. He draws two balls from the urn without replacement. Let X be the number of red balls drawn and Y the number of white balls.

a) Determine the joint probability mass function of X and Y.
b) Are X and Y independent random variables?
c) Compute the covariance between X and Y.

For point A:
$P(0,0)=\frac{4}{9} \cdot \frac{3}{8} = \frac{1}{6}$ That is correct for the solution.
$P(0,1)=\frac{4}{9} \cdot \frac{3}{8} + \frac{3}{9} \cdot \frac{4}{8} = \frac{1}{3}$ That is correct for the solution.
$P(1,0)=\frac{2}{9} \cdot \frac{4}{8} + \frac{4}{9} \cdot \frac{2}{8}= \frac{2}{9}$ That is correct for the solution.
$P(1,1)=\frac{2}{9} \cdot \frac{3}{8} + \frac{3}{9} \cdot \frac{2}{8}= \frac{1}{6}$ That is correct for the solution.
$P(2,0)=\frac{2}{9} \cdot \frac{1}{8} = \frac{1}{36}$ That is correct for the solution.
$P(0,2)=\frac{3}{9} \cdot \frac{2}{8} = \frac{1}{12}$ That is correct for the solution.

For point B to check the independancy I have just to check if for example $P(X=0,Y=0) = P(X=0) \cdot P(Y=0)$.
$\frac{1}{6} \neq (\frac{7}{9} \cdot \frac{6}{8}) \cdot (\frac{6}{9} \cdot \frac{5}{8})$.
So X and Y are not independent.

For point C I know that the covariance $Cov(X,Y)=E[X \cdot Y]-E[X]\cdot E[Y]$, but how can compute the expectations, do I have to figure put with distribution is? How can I do it?

Can someone help me? Thanks in advance, Fabio!

TFAE
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3 Answers3

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For a) keep in mind that the order you draw the balls in matters. For example, for $P(0,1)$ we can draw a blue ball then a white ball or a white ball then a blue ball.

For b) independence says that for every $x$ and $y$, $P(x,y) = P_X(x)P_Y(y)$. It's not enough to check a single pair $(x,y)$ to prove independence. On the other hand, if there is some $x$ and $y$ for which $P(x,y) \ne P_X(x)P_Y(y)$ then $X$ and $Y$ are not independent.

For c) recall that $\displaystyle \mathbf E[f(X,Y)] = \sum_{(x,y)} f(x,y)P(x,y)$.

Trevor Gunn
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  • Thank you for the answer, I understood how to compute correctly the first part, for part B I tried the formula what you wrote, what I did it's correct? For part C how can I compute f(x,y) I have to figure out the distribution type? – TFAE Oct 23 '18 at 22:11
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    @FabioTaccaliti b) looks right. For c) in $\mathbf{E}[XY]$ we have $f(x,y) = xy$ and for $\mathbf{E}[X]$ we have $f(x,y) = x$ and similarly for $Y$. – Trevor Gunn Oct 24 '18 at 00:58
  • And then I have to multiply these $f$ with the probabilities that I calculate in the first part of the problem? – TFAE Oct 25 '18 at 17:53
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    @Fabio Taccaliti yes – Trevor Gunn Oct 25 '18 at 20:42
  • So at the end we should have: $Cov(X,Y)= 1 \cdot 1 \cdot P(1,1) - (1 \cdot P(1,0) + 1 \cdot P(1,1) + 2 \cdot P(2,0) + 1 \cdot P(0,1) + 1 \cdot P(1,1) + 2 \cdot P(0,2))$ – TFAE Oct 25 '18 at 20:49
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    @FabioTaccaliti All the pieces are there correctly but just remember to multiply E[X] by E[Y] rather than add them. – Trevor Gunn Oct 25 '18 at 21:10
  • You are right, thank you for the help! – TFAE Oct 25 '18 at 21:14
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Another way to check dependence (i.e. part B only) is that if $X,Y$ are independent then for any values $x,y$ we have $P(Y=y) = P(Y=y | X=x)$. However, clearly if there are 2 red balls then there cannot be any white balls, i.e. $P(Y > 0) > 0$ but $P(Y > 0 | X=2) = 0$. This might be a more intuitive example demonstrating dependence.

antkam
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but how can compute the expectations, do I have to figure put with distribution is?

You have the distribution, $P(x,y)$. Use it.

The expectation of function $g$ of $X,Y$ is:

$$\begin{align}\mathsf E(g(X,Y)) &=\sum_{x}\sum_{y} g(x,y)~P(x,y)\\[1ex]\textsf{so...}\\[2ex]\mathsf E(X) &=\sum_{x}\sum_{y} x~P(x,y) \\ &=0+P(1,0)+P(1,1)+2P(2,0)\end{align}$$

And so on.

Graham Kemp
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  • Sorry, I don't understand your expression, so, for example, I'll have:
    $E[X] =0 \cdot P(X=0) + 1 \cdot P(X=1) + 2 \cdot P(X=2)$
    $E[Y] =0 \cdot P(Y=0) + 1 \cdot P(Y=1) + 2 \cdot P(Y=2)$
    $E[X,Y] = 0 \cdot 0 \cdot P(X=0,Y=0) + 1 \cdot 0 \cdot P(X=1,Y=0) + 0 \cdot 1 \cdot P(X=0,Y=1) + 1 \cdot 1 \cdot P(X=1,Y=1) + 2 \cdot 0 \cdot P(X=2,Y=0) + 0 \cdot 2 \cdot P(X=0,Y=2)$ right?
    For $P(X=0)$ should I consider just a ball, so $\frac{7}{9}$ or two balls $\frac{7}{9} \cdot \frac{6}{8}$?     – TFAE Oct 23 '18 at 23:37
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    @FabioTaccaliti $\begin{align}P(X{=}0)&=P(X{=}0,Y{=}0)+P(X{=}0, Y{=}1)+P(X{=}0, Y{=}2)\[2ex]&= \sum_y P(X{=}0,Y{=}y)\end{align}$ – Graham Kemp Oct 23 '18 at 23:46
  • I still don't get it, sorry.. – TFAE Oct 25 '18 at 17:57