I am trying to solve the following problem:

Let $X_1,\dots, X_n$, where $n > 4$, be independent random variables such that $X_i ∼ N(i, i)$ for $i = 1, \dots, n$. Let $\bar{X} = {\frac{1}{n}}{\sum} X_i$ be the sample mean.

Calculate the probability that $X_n$ is not the largest observation in the sample. This probability should be expressed in terms of $\Phi (\cdot)$, the cdf of the standard normal distribution.

I am thinking we need to find $P(X_n\leq X_N )\,where\ X_N $ is largest observation. $P(X_n\leq X_N )=P(X_1\leq X_n,X_2\leq X_n,........(X_n\leq X_N)$

$ \hspace{2.7cm} =(P(X_1\leq X_n)P(X_2\leq X_n)........P(X_n\leq X_N))$

But then i lost the path. what should I do? And I am also not sure about my calculation. Any type of suggestion will be appreciated.

  • 1
    $\begingroup$ What do the commas in the last line denote? $\endgroup$ – joriki May 21 '16 at 8:50
  • $\begingroup$ sorry, there should not be any comma. it was a typo. $\endgroup$ – anova May 21 '16 at 15:51
  • $\begingroup$ First, you are after $P(X_n<X_N)$, not after $P(X_n\leqslant X_N)=1$. Second, $P(X_n\geqslant X_N)$ is $P(X_1<X_n,\ldots,X_{n-1}< X_n)$ and $P(X_1<X_n,\ldots,X_{n-1}<X_n)$ is not $P(X_1< X_n)\cdots P(X_{n-1}< X_n)$ since the events $\{X_i<X_n\}$ are not independent. Rather, considering the CDF $F_i$ of each $X_i$, $P(X_1<X_n,\ldots,X_{n-1}< X_n)$ equals $E(F_1(X_n)\cdots F_{n-1}(X_n))$. One can go a little further, but no real simplification occurs after that... $\endgroup$ – Did May 21 '16 at 19:37
  • $\begingroup$ Thanks. So, is there any other way I can try......? $\endgroup$ – anova May 21 '16 at 19:41
  • $\begingroup$ With the varying means and variances that your question specifies, my guess is that one cannot go much farther than what I explain in my previous comment. $\endgroup$ – Did May 21 '16 at 20:25

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