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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 (ΞΈ, Ξ») 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.

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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.

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  • $\begingroup$ Thank you for your help so far!! I plugged everything back in and factorised to get $$𝔼[π‘‰π‘Žπ‘Ÿ[π‘Œ|𝑇]]=𝑇(2πœ†+1)(-πœ†βˆ’1)$$ How do I then get the MoM in terms of (πœƒ,πœ†) ? $\endgroup$
    – user828643
    Oct 13, 2020 at 22:22
  • $\begingroup$ 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 $\endgroup$
    – user828643
    Oct 13, 2020 at 22:24

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