# Method of Moments Estimator of a Compound Poisson Distribution

For the random variable Y constructed as follows:

$$Y = \sum_{i=1}^{T} X_i \$$ where $$T$$~Poisson$$(\lambda)$$ with $$\lambda > 0,\space$$ and$$\space$$ {$${{X_i}}$$}$$^T_{i=1}$$ is an independent and identically distributed sample of size T from a Poisson distribution with mean $$\theta$$.

I have calculated the method of moments estimator for $${\hat\theta}$$ when $$\lambda$$ is known to be $$\frac{\bar{Y}}{{\lambda}}$$.

I now need to derive a method of moments estimator for (θ, λ) based on the sample mean and variance assuming $$\lambda$$ is unknown.

I understand that I need to use the law of total variance however I'm not really sure what to do.

First we use that the sum of $$k$$ Poisson($$\lambda$$) has distribution Poisson($$k \lambda$$). With this, we get that $$Y|T\sim$$Poisson($$T\lambda$$) and therefore $$\mathbb{E}[Y|T]=T \lambda$$.
Now, let us compute the variance. As you mentioned, we can use the law of total variance $$Var[Y] = Var[\mathbb{E}[Y|T]] + \mathbb{E} [Var[Y|T]].$$ Notice that $$Var[\mathbb{E}[Y|T]] = Var[\lambda T]=\lambda^2 Var[T]= \lambda^2 \theta$$.
Now, for the second part, $$\mathbb{E} \left[Var[Y|T]\right] = \mathbb{E} \left[\mathbb{E}[Y^2|T]\right] - 2 \mathbb{E} \left[\mathbb{E}[Y|T]^2\right]+E[T^2].$$ It should be clear how to compute the second and third terms by using the distribution of $$T$$ and $$\mathbb{E}[Y|T]$$. For the first term, notice that \begin{align*} \mathbb{E}[Y^2|T=t] &= \sum_{i,j=1}^t \mathbb{E} [X_i X_j] \\&= \sum_{1\le i\neq j\le t} \mathbb{E} [X_i]\mathbb {E}[ X_j] +\sum_{i=1}^t \mathbb{E} [X_i^2] \\&= \sum_{1\le i\neq j\le t} \mathbb{E} [X_i]^2 +\sum_{i=1}^t \mathbb{E} [X_i^2] \\&= t(t-1)(\lambda^2) +t (\lambda^2+\lambda), \end{align*} where we used that $$X_i's$$ are i.i.d in the second and third identity. Therefore, we have $$\mathbb{E}[Y^2|T]= T(T-1)\lambda^2 + T(\lambda^2+\lambda)$$. Plugin everything back should give you the result. Let me know if you need further help.
• Thank you for your help so far!! I plugged everything back in and factorised to get $$𝔼[𝑉𝑎𝑟[𝑌|𝑇]]=𝑇(2𝜆+1)(-𝜆−1)$$ How do I then get the MoM in terms of (𝜃,𝜆) ?
• would this then give MoMs of $$\hat{\lambda}=\frac{\bar{Y}}{T}$$ and $$\hat{\theta}=\frac{S^2+T(2\frac{\bar{Y}}{T}+1)(\frac{\bar{Y}}{T}-1)}{(\frac{\bar{Y}}{T})^2}$$ where $S^2$ is the variance and $\bar{Y}$ is the expected value