0
$\begingroup$

Five cards are randomly chosen from a standard deck, one at a time with replacement. Let $X$, $Y$, $Z$ be the numbers of chosen queens, kings, and other cards.

(a) Find the joint PMF of X, Y, Z.

(b) Find the joint PMF of X and Y .

a) The joint PMF of $X$, $Y$ and $Z$ $$ P(X=x,Y=y,Z=z) = \left(\frac{1}{13}\right)^{x+y} \left(\frac{11}{13}\right)^z, $$ where the support is $\{(x,y,z) \in \mathbb{Z} \mid x+y+z = 5 \text{ and } x,y,z \geq 0 \}$.

b) To find the joint PMF of $X$ and $Y$ alone, I marginalize out $Z$.

$$ P(X=x,Y=y) = \left(\frac{1}{13}\right)^{x+y} \sum_{z=0}^{5-x-y} \left(\frac{11}{13}\right)^z = \left(\frac{1}{13}\right)^{x+y} \frac{1-\left(\frac{11}{13}\right)^{6-x-y}}{1-\frac{11}{13}} $$

Is this correct?

$\endgroup$
1
$\begingroup$

You are dealing here with multinomial distribution and 'forgot' the coefficients.

Under $x+y+z=5$ and $x,y,z\in\{0,1,\dots\}$:

$$P\left(X=x,Y=y,Z=z\right)=\frac{5!}{x!y!z!}\left(\frac{1}{13}\right)^{x}\left(\frac{1}{13}\right)^{y}\left(\frac{11}{13}\right)^{z}=\frac{5!}{x!y!z!}\frac{11^{z}}{13^{5}}$$

Under $x+y\leq5$ and $x,y\in\{0,1,\dots\}$:

$$P\left(X=x,Y=y\right)=P\left(X=x,Y=y,Z=5-x-y\right)=\frac{5!}{x!y!\left(5-x-y\right)!}\frac{11^{5-x-y}}{13^{5}}$$

$\endgroup$
  • $\begingroup$ Thank you for your answer. Could you explain why plugging in the constraint $5-x-y$ for $Z$ is 'marginalizing' and why not do that already in the joint PMF, as it must be always true? $\endgroup$ – NoBackingDown Dec 30 '14 at 12:57
  • $\begingroup$ You are asked in a) for the PMF of $X$, $Y$ and $Z$ which is exactly finding an expression for $P(X=x,Y=y,Z=z)$ in the variables $x$, $y$ and $z$. This is achieved. You can indeed substitute $z=5-x-y$ if you like, but what do you win by that? The formula will only loose some of its elegance. Nothing is gained. In b) an expression for $P(X=x,Y=y)$ must be found in the variables $x$, $y$. Now you are somehow forced to take that step after all. This because variable $z$ is not allowed on the RHS here. $\endgroup$ – drhab Dec 30 '14 at 13:12
  • $\begingroup$ Thank's for your explanation. I now have a clearer picture. $\endgroup$ – NoBackingDown Dec 30 '14 at 13:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.