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Let $X =\{X_1,X_2,...,X_n\}^T$ such that $X_i \overset{iid}\sim N\left(\theta,1\right)$.

$Y_i=\begin{cases}1,X_i>0\\ 0,\text{otherwise} \end{cases}$

Let $\psi=P(Y_1 = 1)$

Find the maximum likelihood estimator (MLE) $\hat\psi$ of $\psi$.

I'm having trouble understanding what $Y$ represents here. Is it the mean of $X$? In which case would I use $\frac{1}{n}\sum_{i=1}^nX_i$ to estimate the answer?

DeBARtha
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tanvir ahmed
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1 Answers1

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I'm having trouble understanding what Y represents here.

$$\mathbb{P}[X_i>0]=\mathbb{P}[Z>-\theta]=\mathbb{P}[Z<\theta]=\Phi(\theta)$$

thus

$$Y_i=\mathbb{1}_{X_i>0}$$

and

$$\Psi=\Phi(\theta)$$

$\hat{\Psi}_{ML}$ can be derived using $\hat{\theta}_{ML}$ and its invariance property

Example:

You have the following random sample

$$\{2.95;1.19;2.60;1.10\}$$

your estimation is

$$\hat{\Psi}=\Phi(1.96)=0.975$$

tommik
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  • Thank you for clarifying that. I believe I understand now. Would you mind sharing some educational resources on this topic? – tanvir ahmed May 16 '21 at 05:21
  • @tanvirahmed https://www.fulviofrisone.com/attachments/article/446/Introduction%20to%20the%20theory%20of%20statistics%20by%20MOOD.pdf. Chapter VII par 2.2 – tommik May 16 '21 at 05:23
  • Thank you @tommik – tanvir ahmed May 16 '21 at 05:23
  • @tanvirahmed Just a small addition to tommik's answer: ML estimate of $\Phi(\theta) : \Phi(\theta^) = arg max_{\Phi(\theta)}P(X_1,...,X_N | \Phi(\theta)) = \Phi(arg max_{\theta}P(X_1,...,X_N | \Phi(\theta))) $ . hence enough to find: $\theta^ = arg max_{\theta}P(X_1,...,X_N | \Phi(\theta)) = arg max_{\theta}P(X_1,...,X_N | (\theta)) $ as $\Phi(\theta)$ is invertible. Hence $\Phi(\theta^*) = \Phi(\theta_{ML})$. If the function u r trying to estimate is not invertible function of $\theta$ then it will get complicated. – Balaji sb May 16 '21 at 05:40