Question

Consider a geometric distribution where the random variable is the number of the trial with the first success instead of the number of failures before a success. Develop formulas for the probabilities and expectation for this distribution.

I am confused on how to do this question. Please help me!

Answer 1

@amistre64

Answer 2

@hartnn

Answer 3

The geometric distribution (for number of failures before the first success) has the formula
\[p(X=n)=(1-p)^{n}p~~~~\text{for }n=0,1,2,3,\cdots\]
When \(n=0\), for instance, the number of failures is 0, indicating a success happens on the first trial.

So if \(X\) is the random variable for the number of failures, then \(Y=X+1\) should tell you the number of the trial on which the first success occurs, which means you have the distribution
\[\begin{align*}p(Y=n)&=p(X+1=n)\\
&=p(X=n-1)\\
&=(1-p)^{n-1}p
\end{align*}\]
\(Y\) counts the number of the trial of the first success, which means the minimum value of \(n\) must be 1; you can't have a success on the zero-th trial, that makes no sense. So, the support/domain must be \(n=1,2,3,\cdots\).

Finding the expected value shouldn't give you much trouble.