1
$\begingroup$

Let $X_1, ..., X_n$ be a random sample from the $N(\mu,1)$ distribution. First show that $$E \left[ u(X_1)\, \Big|\, \sum^n_{i=1}X_i=z \right] = \Phi \left( \frac{\sqrt n(c-z/n)}{\sqrt{n-1}} \right)\,,$$ where $u(x)=1$ if $x\leq c$ and $0$ otherwise, and $\Phi(x)=P(Z\leq x)$ where $Z$ is $N(0,1)$ distributed. Then secondly determine the MVUE for $\Phi(c-\mu)$ and compare with the MLE of $\Phi(c-\mu)$ for large $n$.

Please could someone point me in the right direction on this question? I'm not sure where to start on the first part. There is a hint in the question that integration by substitution might help, but I can't even see where integration comes into it. Are there any similar questions out there that you could point me to?

For the second part, I know that a MVUE must attain the Cramer Rao lower bound, but because it's a follow on question from the first part I'm pretty stuck on this as well.

Please could you help get me going or explain to me the process that I need to follow. I'm not looking for someone to tell me the answer - just give me a push to help me get going!

(It may not be relevant, but this question is just a bit from a much bigger question. I've already successfully shown that the MLE of $\mu$ is just the sample mean).

Improve this question
$\endgroup$
6
  • $\begingroup$ Please see the answers at stats.stackexchange.com/search?q=expectation+truncated+normal. $\endgroup$
    – whuber ♦
    Commented Jan 9, 2020 at 14:28
  • 1
    $\begingroup$ The lectures haven't covered truncated normals. Are you sure this is relevant? $\endgroup$
    – StatisticsPersonInTraining
    Commented Jan 9, 2020 at 14:42
  • $\begingroup$ Yes--although certainly one can ignore the connection and still solve the problem. Regardless, if you look at some of the related answers you will see how this comes down to a straightforward integration, which is what the hint in the question is getting at. $\endgroup$
    – whuber ♦
    Commented Jan 9, 2020 at 14:44
  • 1
    $\begingroup$ See stats.stackexchange.com/questions/413264/…. $\endgroup$
    – StubbornAtom
    Commented Jan 9, 2020 at 15:10
  • 1
    $\begingroup$ @Jarle This probability is the normalizing constant for a truncated Normal. The purpose of referencing the answers about truncated Normal expectations was to show what the hint in the question might have been referring to. $\endgroup$
    – whuber ♦
    Commented Jan 9, 2020 at 15:20

0

Reset to default