If $X_1, \ldots, X_n \sim N(\theta, 1)$ are iid, I want to see what the MLE of $P(X>0)$ is. One thing that comes to mind is that $\hat{\theta} = \bar{X}$ is the regular MLE of the parameter. By invariance of the MLE, I can use the plug in approach to $\Phi(x) = P(X \leq x)$ by way of $1-\Phi(x) = P(X \geq x)$. However, this results in $1-\Phi(0)$ am so I'm not sure what to do here.
If $X_1, \ldots, X_n \sim N(\theta, 1)$, what is the MLE of $P(X>0)$?
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probability
probability-theory
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0I don't understand why $1-\Phi(x) = P(X\ge x)$ That's true for a standard normal. Maybe you want $1-\Phi(x-\hat\theta)$? – 2017-01-03
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0@spaceisdarkgreen I think you're absolute right, I was told the right answer was that the MLE of $\Phi(\theta)$ is $\Phi(\bar{X})$. Is that correct? – 2017-01-04
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1Looks right to me. – 2017-01-04
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0@spaceisdarkgreen Thanks so much, I totally get it now! – 2017-01-04
1 Answers
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Denote with $\hat{P}(\cdot)$ the MLE of $P(\cdot)$, then \begin{align}\hat P(X>0)&=1-\hat P(X\le 0)=1-\hat P\left(\frac{X-\theta}{1}\le \frac{0-\theta}1\right)\\[0.2cm]&=1-\hat \Phi(-\theta)=^{\text{symmetry of $\Phi$}}\\[0.2cm]&=\hat \Phi(θ)=^{\text{Invariance of the MLE}}\\[0.2cm]&=\Phi(\hat \theta)=\Phi (\bar X)\end{align}