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?

Short 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-by-step solution

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

Explanation (part a)

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

Thus, localid="1647244243081" $P(N=n)=p_0^{n-1}·\left(1-p_0\right)$

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.

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

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.

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

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

localid="1647243142102" $P(X=j,N=n)=P(X=j)·P(N=n)$

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

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

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.

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

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.