Question

Show that the moment generating function of the negative binomial distribution is \(M(t)=p^{r}\left[1-(1-p) e^{t}\right]^{-r}\). Find the mean and the variance of this distribution. Hint: In the summation representing \(M(t)\), make use of the negative binomial series. \({ }^{1}\)

Short answer

The moment generating function of the negative binomial distribution is \(M(t) = p^r \left[ 1 - (1 - p) e^t \right]^{-r}\), the mean is \(\mu = r \cdot \frac{p}{1 - p}\), and the variance is \(\sigma^2 = r \cdot \frac{p}{(1 - p)^2}\).

Step-by-step solution

Prove the Moment Generating Function (MGF) of the Negative Binomial Distribution

This will involve using the power series expression for the moment generating function. Given that the moment generating function \(M(t)\) is defined as the expected value of \(e^{tx}\) where \(x\) is a random variable, we therefore write \[M(t) = \mathbb{E}(e^{tx}) = \sum \^{+\infty}_{k=r} \({ }^{k - 1}\) \( C_{r - 1}\) \( p^r \) \((1 - p)^{k - r}\) \( e^{tk} \)Next, we make use of the negative binomial series \((1 - x)^{-n} = \sum \^{+\infty}_{k=0}\) \({ }^{k + n - 1}\) \( C_{n - 1}\) \( x^k \) for \(|x| < 1\) Hence, we can express \(M(t)\) as\[M(t) = p^r \sum \^{+\infty}_{k=0}\) \({ }^{k + r - 1}\) \( C_{r - 1}\) \([ (1 - p) e^t ]^k \)= p^r \left[ 1 - (1 - p) e^t \right]^{-r} \]

Find the Mean from the MGF

The mean \(\mu\) of a distribution is the first derivative of the moment generating function evaluated at \(t = 0\). This gives:\[\mu = M'(0) = e^t r p^{r - 1} [1 - (1 - p) e^t ]^{-(r + 1)} e^t (1 - p) |_{t=0} = r \cdot \frac{p}{1 - p}\]

Find the Variance from the MGF

The variance \(\sigma^2\) of a distribution is the second derivative of the moment generating function evaluated at \(t = 0\), minus the square of the mean. That is:\[\sigma^2 = M''(0) - [M'(0)]^2 = e^t r p^{r - 1} [1 - (1 - p) e^t]^{-(r + 2)} (1 - p)^2 e^{2t} + e^t r (r + 1) p^{r - 2} [1 - (1 - p) e^t]^{-(r + 2)} e^t (1 - p) |_{t=0} - \mu^2 = r \cdot \frac{p}{(1 - p)^2}\]