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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 $P(X_1=x_1,\ldots,X_n=x_n)$?

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    $\begingroup$ 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? $\endgroup$ – Dilip Sarwate Mar 14 '14 at 15:57
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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.

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I'm treating this question as though it has a self-study tag and giving a suggested method:

  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?

  2. Go from spotting a pattern to a formula. Looking at existing probability distributions is almost always a help in trying to model new ones. There is a list of discrete probability distributions here, and I would suggest starting with the geometric distribution.

  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.

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