Q. 6.38 - A First Course in Probability

Question

Choose a number X at random from the set of numbers $\left\{1,2,3,4,5\right\}$. Now choose a number at random from the subset no larger than X, that is, from $\left\{1 ... ,X\right\}$. Call this second number Y.

(a) Find the joint mass function of X and Y.

(b) Find the conditional mass function of X given that Y = i. Do it for i = $1,2,3,4,5$.

(c) Are X and Y independent? Why? 

Step-by-Step Solution

Answer

(a) The joint mass function of X and Y:

For $k\le l:$

$P\left(Y=k, X=l\right)=\frac{1}{5l}$

For $k>l:$

$P\left(Y=k, X=l\right)=0$

(b) Conditional mass function:

$P(X=l\mid Y=i)=\frac{\frac{1}{5l}}{{\sum }_{i=1}^5\frac{1}{5l}}$

(c) The random variables X and Y are not independent.

Step 1: Given information (part a)

Number X to be chosen at random from $\left\{1,2,3,4,5\right\}$

choose the second number at random from the subset $\left\{1,\dots , X\right\}$

Step 2: Explanation (part a)

We have that $X~D Unif \left(1,...,5\right)$

Observe that $Y\le X$almost certainly.

Also, $Y\in \left\{1,...,5\right\}$

For $k\le l$,

we have 

$P(X=l\mid Y=i)=\frac{P(Y=i,X=l)}{P(Y=i)}=\frac{\frac{1}{5l}}{{\sum }_{i=1}^5\frac{1}{5l}}$

For $k>l:$

$P\left(Y=k,X=l\right)=0$

Step 1: Given information (part b)

Number X to be chosen at random from $\left\{1,2,3,4,5\right\}$

Choose the second number at random from the subset $\left\{1,\dots , X\right\}$

also, $Y=i$ where, $i=1,2,3,4,5$

Step 2: Explanation (part b)

Using total probability law,

$P(Y=i)=\sum _{l=i}^{5}P(Y=i,X=l)=\sum _{l=i}^5\frac{1}{5l}$

then for $i\le l,$

We have $P(X=l\mid Y=i)=\frac{P(Y=i,X=l)}{P(Y=i)}=\frac{\frac{1}{5l}}{{\sum }_{i=1}^5\frac{1}{5l}}$

Step 1: Given information (part c)

Number X to be chosen at random from $\left\{1,2,3,4,5\right\}$

Choose the second number at random from the subset $\left\{1,\dots , X\right\}$

Step 2: Explanation (part c)

Since $Y\le X$,

Then take any $k>l$.

Suppose that take $k=2, l=1$

thus $P\left(Y=k, X=l\right)=0$

On the other hand, it is quite obvious that $P\left(Y=k\right)>0$ and $P\left(X=l\right)>0$.

Thus, $P(Y=k,X=l)\ne P(Y=k)P(X=l)$

Therefore, they are not independent