Q. 6.24 - A First Course in Probability

Question

Consider independent trials, each of which results in outcome i, i = $0,1, ....,k$, with probability $p_i,\sum _{i=0}^K{p}_i=1$ . Let N denote the number of trials needed to obtain an outcome that is not equal to $0$, and let X be that outcome.

(a) Find $P\{N=n\},n\ge 1$ 

(b) Find  $P\{X=j\},j=1,\dots ,k$

(c) Show that $P\{N=n,X=j\}=P\{N=n\}P\{X=j\}$.

(d) Is it intuitive to you that N is independent of X? 

(e) Is it intuitive to you that X is independent of N? 

Step-by-Step Solution

Answer

(a) $P(N=n)=p_0^{n-1}·\left(1-p_0\right)$

(b) $P(X=j)=\frac{p_j}{1-p_0}$

(c) X and N are independent

(d) Yes, it is intuitive that N is independent of X

Step 1: Given information (part a)

For independent trials, each resulting in the outcome,

$i,i=0,1, ...,k$

with probability

$p_i,\sum _{i=0}^k{p}_i=1$

N: number of trials needed to obtain an outcome.

X: Outcome obtained

Step 2: Explanation (part a)

N has Geometric distribution with a parameter of success, $1-p_0$

Thus, $P(N=n)=p_0^{n-1}·\left(1-p_0\right)$

This is due to the fact that

if $N=n$

That implies in the first $n-1$trails, we have obtained outcomes equal to zero and in n^th trial, we have obtained any other outcome rather than zero.

Step 1: Given information (part b)

For independent trials, each resulting in the outcome,

$i,i=0,1, ...,k$

with probability

$p_i,\sum _{i=0}^k{p}_i=1$

N: number of trials needed to obtain an outcome.

X: Outcome obtained

Step 2: Explanation (part b)

What so ever be the time of the first non-zero outcomes, every other no zero has the proportional probability to be that outcome. 

such that

$P(X=j)=\frac{p_j}{1-p_0}$

Now observe that we divide p_j with $1-p_0$.

Because we know that considered outcome cannot be zero and the sum of all probability should be equal to one.

Step 1: Given information (part c)

For independent trials, each resulting in the outcome,

$i,i=0,1, ...,k$

with probability

$p_i,\sum _{i=0}^k{p}_i=1$

N: number of trials needed to obtain an outcome.

X: Outcome obtained

Step 2: Explanation (part c)

We have

$P(X=j,N=n)=P(X=j\mid N=n)·P(N=n)$

Since the time of first non-zero outcomes doesn't matter,

then we have 

$P(X=j\mid N=n)=P(X=j)$

Thus

$P(X=j,N=n)=P(X=j)·P(N=n)$

Hence we have proved that X and N are independent and the joint probability is the product of individual probability.

Step 1: Given introduction (part d)

for independent trials, each resulting in outcome,

$i,i=0,1, ...,k$

with probability

$p_i,\sum _{i=0}^k{p}_i=1$

N: number of trials needed to obtain an outcome.

X: Outcome obtained

Step 2: Explanation (part d)

We have shown that

$P(X=j,N=n)=P(X=j)·P(N=n)$

Now, when the first non-zero outcomes are known, that does not imply when the first non-zero outcomes might happen. 

Therefore N is independent of X.

Step 1: Given information (part e)

For independent trials, each resulting in the outcome,

$i,i=0,1, ...,k$

with probability

$p_i,\sum _{i=0}^k{p}_i=1$

N: number of trials needed to obtain an outcome.

X: Outcome obtained

Step 2: Explanation (part e)

In part (c) we have shown that

$P(X=j,N=n)=P(X=j)·P(N=n)$

Now, when the happening of first non-zero outcomes is known

that does not tell the value of that first non-zero outcome. Therefore, X is independent of N.