# Independent variables, normal distribution, pdf

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 $Z_i$

• Do you mean uniform distribution? The normal distribution isn't bounded. – Nick Peterson May 26 '14 at 14:31
• my mistake, it should be unbounded – user3676690 May 26 '14 at 14:50
• @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? – Michael Hardy May 26 '14 at 15:44
• 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. – Michael Hardy May 26 '14 at 15:47

## 3 Answers

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, $$f_W(w)=n\lambda\mathrm e^{-\lambda w}(1-\mathrm e^{-\lambda w})^{n-1}\mathbf 1_{w\gt0}.$$

• why $P(\forall i,Z_i\lt w)=P(Z_1\lt w)^n. $$– user3676690 May 26 '14 at 15:28 • Thank you, I have to go, but I'll check it and answer in 2-3 hours – user3676690 May 26 '14 at 15:31 If Z = \varphi(X), then X = \varphi^{-1}(Z) = 1-e^{- \lambda Z} and \frac{dX}{dz} = 1 +\lambda e^{- \lambda z}. Now use the definition of the function of random variables. • I don't understand why you calculate X . How it help to find pdf of Z – user3676690 May 26 '14 at 14:53 • en.wikipedia.org/wiki/… – Alex May 26 '14 at 14:54 • I updated my question. Now I have pdf of each Z_i , but each is given by the same formula, so how to get maximum ? – user3676690 May 26 '14 at 15:06 • what maximum are you talking about? you wanted pdf of a function of rv – Alex May 26 '14 at 15:13 • sorry, I didn't write all in my question. I updated it again – user3676690 May 26 '14 at 15:16 It appears that you meant the uniform distribution on [0,1]. If 0<X<1 then Z=-\ln(1-X)>0. Then we have$$ \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.