Question

Consider a sequence of independent Bernoulli trials, each of which is a success with probability p. Let X1 be the number of failures preceding the first success, and let X2 be the number of failures between the first two successes. Find the joint mass function of X1 and X2.

Answer 1

Answer:

$P(X_1,X_2) = \left \{ {{p^2(1-p)^{k+l}, \quad k,l = 0,1,2,\ldots} \atop {0, \quad \text{otherwise}}} \right.$

Step-by-step explanation:

Let $X_1$ be a random variable that counts the number of failures preceding the first success and $X_2$ be a random variable that counts the  number of failures between the first two successes. The probability of success in a independent trial is $p$ and both of them have Bernoulli distribution.

Assume that the number of unsuccessful trials before the first successful one is $k$ and that $l$ is a number of failures between two successes.

The joint mass function of two discrete random variables $X$ and $Y$ is defined by

                                $P_{XY}(x,y) = P(X=x,Y=y)$

In this case, we have two discrete random variables $X_1$ and $X_2$ which are independent, so we have that their joint mass function is of the form

              $P_{X_1,X_2}(x,y) = P(X_1=x,X_2=y) = P(X_1=x) \cdot P(X_2=y)$

The probability that the number of failures before the first successful trial is equal to $k$ is

                     $P(X_1=k) = \underbrace{(1-p)^k}_{\text{in k trials a failure}} \cdot \underbrace{p}_{\text{after that,a success}}$

The probability that the number of failures between two successes is equal to $l$ is

                          $P(X_2 = l) = \underbrace{p}_{\text{a success}} \cdot \underbrace{(1-p)^{j}}_{\text{ j failures}} = p(1-p)^j$

Therefore,

                           $P(X_1,X_2)= P(X_1=k) \cdot P(X_2=l) \\\phantom{P_{X_1,X_2} \; =} = (1-p)^k \cdot p \cdot p (1-p)^l \\\phantom{P_{X_1,X_2} \; =} = (1-p)^{k+l} \cdot p^2$

Now, we obtain that that their joint mass function is

                           $P(X_1,X_2) = \left \{ {{p^2(1-p)^{k+l}, \quad k,l = 0,1,2,\ldots} \atop {0, \quad \text{otherwise}}} \right.$