Independent variables, normal distribution, pdf

Question

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 $

Answer 1

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

Answer 2

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.

Answer 3

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.