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Let be $(X_1,Y_1),...,(X_n,Y_n)$ n independent couples of random variables where $X_i\sim{Binom(10,\theta)}$ and $Y_i|X_i=x_i\sim{Pois(\lambda{(1+x_i)})}$. I found the MLE for $\lambda$, which is: $$ \hat{\lambda}={\sum_{i=1}^n{Y_i}\over{n+\sum_{i=1}^n{X_i}}}. $$ How can I prove that it is a consistent estimator for $\lambda$?

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I will let you fill in the remaining details. $$\hat\lambda = \dfrac{\sum_{i=1}^{n}Y_i}{n + \sum_{i=1}^{n}X_i} = \dfrac{\frac{1}{n}\sum_{i=1}^{n}Y_i}{\frac{1}{n}(n + \sum_{i=1}^{n}X_i)} = \dfrac{\frac{1}{n}\sum_{i=1}^{n}Y_i}{1 + \frac{1}{n}\sum_{i=1}^{n}X_i}\text{.}$$ We observe $$\mathbb{E}[Y_i] = \mathbb{E}[\mathbb{E}[Y_i \mid X_i]] = \mathbb{E}[\lambda(1+X_i)] = \lambda(1+10\theta)\text{.}$$ Now observe $$\frac{1}{n}\sum_{i=1}^{n}X_i \to 10\theta$$ in probability (do you see why?), so by Mann-Wald (or continuous mapping), the function $g: \mathbb{R}_{\geq 0} \to \mathbb{R}$ given by $$g(x) = \dfrac{1}{1+x}$$ is continuous, hence $$\dfrac{1}{1 + \frac{1}{n}\sum_{i=1}^{n}X_i} \to \dfrac{1}{1+10\theta}$$ hence (why?) $$\hat\lambda \to \lambda$$ in probability, as desired.

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