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Given discrete random variables $X_1, \ldots X_n$ with probability mass function: $$ f(x;\theta) = \begin{cases} \theta & x= -1 \\ (1-\theta)^2 \theta^x & x = 0, 1, \ldots \\ \end{cases} $$

Prove that the MLE of $\theta$ is: $$ \hat{\theta}_n = \frac{2\sum_{i=1}^{n}I_{(X_i=-1)} + \sum_{i=1}^{n}X_i}{2n + \sum_{i=1}^{n}X_i} $$

My attempt is as follows:

Let $Y = \mathbb{1}_{[X_i=-1]}$. The likelihood function can be set up as: $$ L(\theta|x) = \Pi_{i=1}^{n}f(x_i;\theta)= \theta^{\sum_{i=1}^{n}Y}\cdot\theta^{\sum_{i=1}^{n}\left((1-Y)X_i\right)}\cdot(1-\theta)^{2\sum_{i=1}^{n}(1-Y)} $$

Then the log-likelihood is: $$ \begin{align} l(\theta|x) &= \log\Pi_{i=1}^{n}f(x_i;\theta) = \log(\theta^{\sum_{i=1}^{n}Y}\cdot\theta^{\sum_{i=1}^{n}\left((1-Y)X_i\right)}\cdot(1-\theta)^{2\sum_{i=1}^{n}(1-Y)})\\ &= \log(\theta) \sum_{i=1}^{n}Y + \log(\theta)\sum_{i=1}^{n}\left((1-Y)X_i\right) + \log(1-\theta)\cdot2\sum_{i=1}^{n}(1-Y) \end{align} $$

Solving by direct maximization, $\frac{d}{d\theta}l(\theta|x) = 0$: $$ \frac{\sum_{i=1}^{n}Y}{\theta} + \frac{\sum_{i=1}^{n}(1-Y)X_i}{\theta} - \frac{2\sum_{i=1}^{n}(1-Y)}{1-\theta} = 0 $$

By solving at the end I arrive at my MLE candidate being: $$ \hat{\theta}_n = \frac{\sum_{i=1}^{n}\mathbb{1}_{[X_i=-1]} + \sum_{i=1}^{n}X_i - \sum_{i=1}^{n}\mathbb{1}_{[X_i=-1]}X_i}{2n+\sum_{i=1}^{n}X_i - \sum_{i=1}^{n}\mathbb{1}_{[X_i=-1]} - \sum_{i=1}^{n}\mathbb{1}_{[X_i=-1]}X_i} $$

Which is close to the solution (I know I still should check if it's the global maximum). I can't see if I am making any algebraic mistake or the likelihood function is not correctly defined. Any help or hint appreciated!

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I don't think you have made any mistakes, and you are very close to the solution.

Note that $\sum_{i=1}^{n}\mathbb{1}_{[X_i=-1]}X_i=-\sum_{i=1}^{n}\mathbb{1}_{[X_i=-1]}$ - substituting this in your equation will result in the solution.

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