# Finding the joint probability distribution $P(X_1=x_1,\ldots,X_n=x_n)$

Let $Y_1,Y_2,\ldots$ be a sequence of independent Bernoulli trials with parameter $p$ and $X_1,X_2,\ldots$ be respectively the first time of success, second time of success,$\ldots$. How can I calculate the joint probability distribution $P(X_1=x_1,\ldots,X_n=x_n)$?

• Another one of hadisanji's homework problems, showing no personal attempts at solving whatsoever, to which we are asked to provide complete answers that will rarely ever be accepted? – Dilip Sarwate Mar 14 '14 at 15:57

Given $x_1<x_2<\dots<x_n$, use the product rule $$P(X_1=x_1,\dots,X_n=x_n) = P(X_1=x_1)\times P(X_2=x_2\mid X_1=x_1)\times\dots\times P(X_n=x_n\mid X_{n-1}=x_{n-1},\dots, X_1=x_1) \, .$$ In this problem we have a markovian property $$P(X_k=x_k\mid X_{k-1}=x_{k-1},\dots, X_1=x_1) = P(X_k=x_k\mid X_{k-1}=x_{k-1}) = P(Z=x_k-x_{k-1}) \, ,$$ in which $Z$ is a negative-binomial random variable with probability of success equal to $p$. Put everything together to find the joint distribution.
1. Write out some examples. E.g. $P(X_1=K_1=2, X_2=k_2=5, X_3=k_3=9)$ is the probability of getting $010010001$. See if you can spot a pattern, e.g. $010010001$ is 1 failure followed by 1 success followed by 2 failures followed by 1 success followed by 3 failures followed by 1 success. Can you relate this to the values of $X_1$, $X_2$, and $X_3$? Can you generalize?
3. Have a look at your proposed formula. Can you simplify it? Are there any particular values of $X_1,\ldots,X_n$ for which it wouldn't work and for which you need to make exceptions? Finally, test it against simple examples.