$X_1,\ldots,X_n,\; n\ge1$ — independent random variables $U(0,1),$

$S_n=\sum_{i=1}^n X_i,$





$\mathrm{cov}(S_n,Z_n)=\mathrm E(S_nZ_n)-\mathrm ES_n \mathrm EZ_n,$

$\mathrm E(S_nZ_n)=n \mathrm E(X_1 \max(X_1,\ldots,X_n)).$

Let $U=\max(X_1,\ldots,X_n),\,V=X_1$.

For $0 \le v \le u \le 1$ we have

$ F(v,u) = \mathbb P(V \le v, U \le u) = \mathbb P(U \le u) - \mathbb P(U \le u, V > v) = \mathbb P(X_1 \le u,\ldots, X_n \le u) - \mathbb P(v < X_1 \le u, X_2 \le u,\ldots, X_n \le u) = vu^{n-1} .$

One don't need to calculate the distribution function for the other areas.

$f(v,u)= \frac{\partial^2}{\partial v \, \partial u} F(v,u) = (n-1)u^{n-2},$

$ \mathrm E(VU) = \int_0^1 \int_0^u vu(n-1)u^{n-2} \mathrm dv \, \mathrm du = \frac{n-1}{2(n+2)},$

$ \mathrm E(S_nZ_n) = \frac{n(n-1)}{2(n+2)} ,$

$ \mathrm ES_n = n \mathrm EV = \frac n2 ,$

$ \mathrm EZ_n = \mathrm EU = \frac n{n+1} .$

What is wrong? The problem must be with $f(v,u)$ because $ \int_0^1 \int_0^u f(v,u) \mathrm dv \, \mathrm du \neq 1 . $

  • 1
    $\begingroup$ Why the downvote? I can´t comprehend that. $\endgroup$ – callculus Jul 13 '18 at 19:57
  • $\begingroup$ One of the $X_k$ is $=Z_n$. You need to treat that one special. $\endgroup$ – herb steinberg Jul 13 '18 at 20:58
  • $\begingroup$ I should have said one of the $X_k$ is $Z_n$, not just equal. $\endgroup$ – herb steinberg Jul 14 '18 at 1:28

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