Probability mass function of max and min function

Question

You repeatedly draw a number at random from the numbers $1,2,...,10$. Let $X$ be the number of drawings until the number $1$ appears and $Y$ be the number of drawings until the number $10$ appears. What is the joint probability mass function of $X$ and $Y$? What are the probability mass functions of $\max (X,Y)$ and $\min (X,Y)$?

Attempt

$X$ and $Y$ are geometric both with parameter $\dfrac{1}{10}$ since each number is equally likely to occur when picking a number at random. We need to find $P(X=x, Y=y)$. Here I get confused as to whether make the assumption that both $X$ are $Y$ are independent. Can we assume this? IF so , then

$$ P(X=x,Y=y) = (0.9)^{x-1} 0.1 (0.9)^{y-1} 0.1 $$

Now, for the next part notice that $max(X,Y)=n$ iff $X \leq n \; \text{or} Y \leq n$.hence,

$$ P(\max(Y,X) = n) = P(X \leq n) + P(Y \leq n) = \sum_{x=1}^n (0.9)^{x-1} 0.1 + \sum_{y=1}^n (0.9)^{y-1} 0.1 $$

Similarly,

$$ P( \min (Y,X) = m) = O(Y \geq n) + (X \geq n) = \sum_{y=n}^{\infty} (0.9)^{y-1} 0.1 + \sum_{ x =n}^{\infty} (0.9)^{x-1} 0.1 $$

Am I correct?

Answer 1

$X$ and $Y$ are not independent since if $P(X=1,Y=1) =0$ but $P(X=1)>0$ and $P(Y>1)>0$.

if $1\le x < y$,

$$Pr(X=x, Y=y) = (0.8)^{x-1}(0.1)(0.9)^{y-x-1}(0.1)=0.8^{x-1}(0.9)^{y-x-1}(0.1)^2$$

That is the first $x-1$ times can be anything besides the two numbers, followed by $1$, then the following $y-x-1$ times we can get any number besides $10$.

Try to work out what happens when $x=y$ and for the case where $x>y$.

Notice that we can't conclude $\max(X,Y)=n \iff X \le n\lor Y \le n$ as if $X=n-2$ and $Y=n-1$, then we ahve $X \le n \lor Y \le n$ being true but $\max(X,Y)=n-1$.

\begin{align} Pr(\max(X,Y)=n) &= Pr(Y > X \land Y=n) + Pr(X > Y \land X=n) \\ &= \sum_{x=1}^{n-1}Pr(X=x,Y=n) + \sum_{y=1}^{n-1}Pr(X=n,Y=y) \end{align}

and

\begin{align} Pr(\min(X,Y)=n) &= Pr(Y > X \land X=n) + Pr(X > Y \land Y=n) \\ &= \sum_{y=n+1}^{\infty}Pr(X=n,Y=y) + \sum_{y=n+1}^{\infty}Pr(X=x,Y=n) \end{align}

Answer 2

$X,Y$ are not independent. If they were, then $P(X,Y=1)$ would be $P(X=1)P(Y=1)=\frac1{100}$, but it is zero instead – you cannot draw two numbers at once. The same reasoning applies for $X=Y$, so $f_{X,Y}(x,y)=0$ if $x=y$.

Otherwise, $f_{X,Y}(x,y)$ where $x<y$ is the probability that the following sequence of draws occurs: $x-1$ times no draw of 1 or 10, draw 1, $y-x-1$ times no draw of 10 (1 can occur again in these draws), draw of 10. This has probability $$\left(\frac45\right)^{x-1}\frac1{10}\left(\frac9{10}\right)^{y-x-1}\frac1{10}=\frac1{100}\left(\frac45\right)^{x-1}\left(\frac9{10}\right)^{y-x-1}$$ Similar reasoning applies to $x>y$, resulting in a joint pmf of $$f_{X,Y}(x,y)=\frac1{100}\left(\frac45\right)^{\min(x,y)-1}\left(\frac9{10}\right)^{|y-x|-1}1_{x\ne y}$$ Now $\min(X,Y)$ is simply the number of draws up to and including the first 1 or 10, so is geometric with success rate $\frac15$. As for $\max(X,Y)$, the probability that it is $n$ is twice the probability that

• the first $n-1$ draws are not 1, but there is at least one 10: $(9/10)^{n-1}-(8/10)^{n-1}$
• the next roll is 1: $\frac1{10}$

Hence $$f_{\max(X,Y)}(n)=\frac15\left(\left(\frac9{10}\right)^{n-1}-\left(\frac45\right)^{n-1}\right)$$