What is the expected value of number of Bernoulli trials required for k successes? Assume probability of success in a single trial = $p$, probability of failure = $q = 1 – p$.
I managed to derive the discrete probability mass function:
$P(X = n) = \binom{n – 1}{k -1} p^{k} (1 – p)^{n – k}$, for $n \geq k$. However, I could not sum the series $E(X) = \sum_{n = k}^{\infty} n\binom{n – 1}{k -1} p^{k} (1 – p)^{n – k}$. Can anybody please help?
Bernoulli Trials Required for k Successes – Probability Distributions
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Best Answer
If you want to analytically calculate the sum, you can do as follows: $$ \sum_{n = k}^{\infty} n\binom{n - 1}{k -1} p^{k} (1 - p)^{n - k}=(\frac{p}{1-p})^k\sum_{m = 0}^{\infty} (m+k)\binom{m+k - 1}{m} (1 - p)^{m+k} $$
Now you can find the following identity in the literature: $$ \sum_{m = 0}^{\infty} \binom{m+k - 1}{m} y^{m}=\frac{1}{(1-y)^k} $$ Now by multiplication of $y^k$ and derivation wrt $y$, you get: $$ \sum_{m = 0}^{\infty} (m+k)\binom{m+k - 1}{m} y^{m+k-1}=\left(\frac{y^k}{(1-y)^k}\right)^{'}=\left(\frac{ky^{k-1}}{(1-y)^{k+1}}\right) $$ Now put $y=1-p$ in the first identity and you get: