Method of moments for an overdispersed count

Question

You count the number of losing trades before your $r$ -th winning trade. This is a negative binomial variable with unknown number of required wins $r$ and unknown win probability $p$ . From your data the sample mean is $2$ and the sample variance is $3$ . (Notice the variance exceeds the mean, which is why a Poisson would not fit.)

Use the method of moments to estimate $r$ and $p$ .

Accepted Answer

The two model moments. Counting failures before the r-th success, the negative binomial has \mathbb{E}[X] = \frac{r(1 - p)}{p}, \qquad \operatorname{Var}(X) = \frac{r(1 - p)}{p^2}. Because p < 1, the variance is always larger than the mean here: that gap is the "overdispersion" a single-rate Poisson cannot produce. Solve for p first. Divide the variance by the mean: \frac{\operatorname{Var}(X)}{\mathbb{E}[X]} = \frac{1}{p} \implies \hat p = \frac{	ext{mean}}{	ext{variance}} = \frac{2}{3} \approx 0.667. Then solve for r. From the mean, r = \dfrac{	ext{mean} \cdot p}{1 - p}, which is the sam…

Method of moments for an overdispersed count

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