Number of failures until first r^th success - Expectation, Variance, Probability

Question

I was given the following problem:

Given:

For the r.v $\displaystyle F=number\ of\ failures\ before\ first\ success$.

$\displaystyle E[ F] \ =\ \frac{1-p}{p}$ and $\displaystyle Var[ F] \ =\ \frac{1-p}{p^{2}}$.

Let's now define $\displaystyle X=number\ of\ failure\ before\ r^{th} \ success$, calculate the $\displaystyle E[ X] ,\ Var[ X]$.

My solution was:

Let's notice that if $\displaystyle X=the\ number\ of\ failures\ till\ r^{th} \ success$, then we can define it using the previous question. We have $\displaystyle r$ successes, each success has an $\displaystyle i$ number of falilures before it, for $\displaystyle 0\leqslant i\leqslant k$.

So we can define:

$\displaystyle X\ =F_{1} +F_{2} +F_{3} +...+F_{r}$.

Where each $\displaystyle F_{i} =the\ number\ of\ failures\ after\ the\ ( i-1)^{th} \ success\ and\ before\ the\ i^{th} \ success$.

For example:

$\displaystyle F_{1}$ will be the number of failures before the $\displaystyle 1^{st}$ success.

$\displaystyle F_{2}$ will be the number of failure between the $\displaystyle 1^{st}$ and $\displaystyle 2^{nd}$ success.

$\displaystyle F_{3}$ will be the number of failure between the $\displaystyle 2^{nd}$ and $\displaystyle 3^{rd}$ success.

etc.

Note that each $\displaystyle F_{i}$ is simply the number of failures before some first success, meaning, it is exactly the random variable $\displaystyle F$ we were given.

Expectation:

$\displaystyle E[ X] =E[ F_{1} +F_{2} +F_{3} +...+F_{r}] =E[ F_{1}] +[ F_{2}] +[ F_{3}] +...+[ F_{r}] =r\cdotp \frac{1-p}{p} .$

Variance:

As per the $\displaystyle Var[ X]$, I'm having some trouble calculating it.

I can't use linearity with Variance, and I fail to understand wether I can simply say that:

$\displaystyle Var[ X] \ =\ Var[ rF] =r^{2} Var[ F]$.

Or, I should prove that each $\displaystyle F_{1} +F_{2} +F_{3} +...+F_{r}$ is indepentent, and so deduct that:

$\displaystyle Var[ X] \ =rVar[ F]$.

Which one of the two is correct? And how would I prove that the $\displaystyle F_{i}$ are independet?

Answer 1

You are correct. This is the negative binomial distribution and if you think a bit, you'll realise that any failure will be independent of the previous succeses that have occurred so the variance would be the $r\frac{1-p}{p^2}$

Answer 2

The $(F_k)$ are a sequence of identically distributed random variables, but they not identical. Thus $\mathsf{Var}(\sum_{k=1}^r F_k)$ does not equal $\mathsf {Var}(k F)$.

So, once you have successfully argued that they are also independent, then you may use:$$\begin{align}\mathsf{Var}(\sum_{k=1}^n F_k)&=\sum_{k=1}^n \mathsf{Var}(F_k)\\[1ex]&=n\,\mathsf{Var}(F)\end{align}$$

And how would I prove that the Fi are independent?

Independently from the count for failures and successes that have preceded any success, the trials after that will still form a sequence of independent Bernoulli random variables with the identical success rate.