More precisely i have the following problem:

given a sample of r.v. $\{X_i\}_{i=1...n}$ i.i.d. distributed with $$f_{\theta}(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(x-\theta)^2}$$ and the statistic $$ T(X_1,...,X_n) = \frac{\sum_{i=1}^{i=n} X_i }{n}$$ for $\theta$, let's consider $$g(\theta) = P(X_1 \leq y) = \int_{- \infty}^{y}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(x-\theta)^2}dx$$ and the function $$u(t) = E[\mathbb{1}_{X_1 \leq y}(X_1,...,X_n)|T = t],$$ where $\mathbb{1}_{X_1 \leq y}(X_1,...,X_n)$ is the indicator function of the event $\{X_1 \leq y\}$.

Question : Evaluate $u(t)$ and verify that $u(t)$ doesn't depend on $\theta$, and evaluate the $Var_{\theta}(u(T(X_1,...,X_n)))$.

Obviously $u(t) = E[\mathbb{1}_{X_1 \leq y}(X_1,...,X_n)|T = t] = P(X_1 \leq y |\sum_{i=1}^{i=n} X_i = nt)$, but how i can evaluate it? I tried to apply the trasformation in Find the joint distribution of $X_1$ and $\sum_{i=1}^n X_i$, that is a similar problem but with a different distribution function, but give me nothing because if i calculate the integral, in order to find the conditional probability, remain the dependency on $\theta$ and the integral is not computable. Probably the approach is different.

I have no idea how to approach it!

Generally, how i can evaluate $u(t)$ when there are continuos distributions? There is a common startegy to evaluate this integral?

  • 1
    $\begingroup$ Here is a hint: $P(X_1\le y|\sum_{i=1^n} X_i=nt)=P(X_1\le y|X_1 = nt - \sum_{i=2}^n X_i)$. Then, what is the result if $nt - \sum_{i=2}^n X_i$ is equal, say, to $y+42$ ? $\endgroup$
    – TMat
    Sep 1, 2020 at 11:30
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    $\begingroup$ Does this answer your question? UMVUE for probability of cutoff $\endgroup$ Sep 1, 2020 at 11:46
  • $\begingroup$ @Xi'an in this case, how can i do that? Because the approach in the link that I mentioned doesn't work, there are too much integrals that are not computable. And why if $\{X_i\}_i$ are indipendent then their sum are also indipendent from $X_1$? $\endgroup$
    – Tazz
    Sep 1, 2020 at 12:49
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    $\begingroup$ You can immediately reduce the problem to one with two variables, $X_1$ and $X_2+\cdots+X_n,$ which have a joint Normal distribution. $\endgroup$
    – whuber
    Sep 1, 2020 at 13:41
  • 1
    $\begingroup$ I am afraid that, if this Normality step is beyond your reach, the homework question may prove too advanced for your probability skills. $\endgroup$
    – Xi'an
    Sep 1, 2020 at 16:31


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