0
$\begingroup$

I have independent variables $ X_1, X_2,\ldots,X_n $ with normal distribution on range $ [0,1] $ . Next, variables $ Z_i $ are created according to this formula $ Z_i = - \frac{1}{\lambda} \ln(1-X_i) $ , where $ \lambda > 0 $. How to find probability density function of $ W = \max (Z_1 , \ldots , Z_n ) $ ?

After comments : So, I have a formula $ f_Z(z) = | \frac{d}{dz} (1-e^{-\lambda z}) | f_X(1-e^{-\lambda z}) $ , but $ f_X (x) = 1 $ , so $ f_Z(z) = 1 + \lambda e^{-\lambda z} $ And it represents all $ Z_i $

$\endgroup$
  • 4
    $\begingroup$ Do you mean uniform distribution? The normal distribution isn't bounded. $\endgroup$ – Nick Peterson May 26 '14 at 14:31
  • $\begingroup$ my mistake, it should be unbounded $\endgroup$ – user3676690 May 26 '14 at 14:50
  • 1
    $\begingroup$ @user3676690 : You're being quite unclear. You say "normal distribution on range $[0,1]$ and then in comments that that's a mistake and it should be unbounded. That leaves the question of which normal distribution it is. Then you have $\ln(1-X_i)$, which does not exist if $X_i>1$, which makes it look as if restricting it to $[0,1]$ makes sense, but then of course it's not normally distributed. Could you edit the post to make it clear what you meant? $\endgroup$ – Michael Hardy May 26 '14 at 15:44
  • $\begingroup$ Later you say $f_X(x)=1$, so apparently you did mean the uniform distribution, so your comment that it should be unbounded makes no sense. $\endgroup$ – Michael Hardy May 26 '14 at 15:47
1
$\begingroup$

For every maximum $W$ of random variables, one has, for every $w$, $$ [W\lt w]=\bigcap_{i=1}^n[Z_i\lt w]. $$ If $(Z_i)_{1\leqslant i\leqslant n}$ is i.i.d. and distributed as $Z$, then, for every $w$, $$ P(W\lt w)=\prod_{i=1}^nP(Z_i\lt w)=P(Z\lt w)^n. $$ In your case, considering $X$ distributed as $X_1$, $$ [Z\lt w]=[\ln(1-X)\gt-\lambda w]=[X\lt1-\mathrm e^{-\lambda w}], $$ hence, for every $w\gt0$, $$ F_W(w)=F_Z(w)^n=F_X(1-\mathrm e^{-\lambda w})^n. $$ If every $X_i$ is uniform on $[0,1]$ then $F_X(x)=x$ for every $x$ in $(0,1)$ hence, for every $w\gt0$, $$ F_W(w)=(1-\mathrm e^{-\lambda w})^n. $$ Then the PDF $f_W$ is obtained by differentiation, that is, $$ f_W(w)=n\lambda\mathrm e^{-\lambda w}(1-\mathrm e^{-\lambda w})^{n-1}\mathbf 1_{w\gt0}. $$

$\endgroup$
  • $\begingroup$ why $ P(\forall i,Z_i\lt w)=P(Z_1\lt w)^n. $$ $ $\endgroup$ – user3676690 May 26 '14 at 15:28
  • $\begingroup$ Thank you, I have to go, but I'll check it and answer in 2-3 hours $\endgroup$ – user3676690 May 26 '14 at 15:31
0
$\begingroup$

If $Z = \varphi(X)$, then $X = \varphi^{-1}(Z) = 1-e^{- \lambda Z}$ and $\frac{dX}{dz} = 1 +\lambda e^{- \lambda z}$. Now use the definition of the function of random variables.

$\endgroup$
  • $\begingroup$ I don't understand why you calculate $ X $ . How it help to find pdf of $ Z $ $\endgroup$ – user3676690 May 26 '14 at 14:53
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – Alex May 26 '14 at 14:54
  • $\begingroup$ I updated my question. Now I have pdf of each $ Z_i $, but each is given by the same formula, so how to get maximum ? $\endgroup$ – user3676690 May 26 '14 at 15:06
  • $\begingroup$ what maximum are you talking about? you wanted pdf of a function of rv $\endgroup$ – Alex May 26 '14 at 15:13
  • $\begingroup$ sorry, I didn't write all in my question. I updated it again $\endgroup$ – user3676690 May 26 '14 at 15:16
0
$\begingroup$

It appears that you meant the uniform distribution on $[0,1]$. If $0<X<1$ then $Z=-\ln(1-X)>0$. Then we have $$ \Pr(Z\le z) = \Pr(-\ln(1-X)\le z) = \Pr(X\le 1-e^{-z}) = 1-e^{-z}. $$ So $$ \Pr(\max\{Z_1,\ldots,Z_n\} \le z ) = \Pr(Z_1\le z\ \&\ \cdots\ \&\ Z_n\le z) $$ $$ = (\Pr(Z_1\le z))^n = (1-e^{-z})^n. $$ Differentiate to get the density function.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.