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Let $X\sim \mathcal{U}(0,1)$ and $Y = \ln\left((1-X)^{-c}\right)$ for some $c>0$.

We need to show that $Y \sim {\mathrm{Exp}}(1/c)$.

We got to here but couldn't go further:

$$ \mathbb{P}(Y \leqslant y) = \mathbb{P}( X \leqslant 1- \mathrm{e}^{-c y}) $$

Thanks in advance.

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Minor glitch, $\ln(1-X)^{-c} \le y$ iff $(1-X)^{-c}\le e^{y}$ iff $1-X)^c \ge e^{-y}$ iff $1-X \ge e^{-y/c}$ iff $X\le 1-e^{-y/c}$. Then differentiate. –  André Nicolas Jan 8 '12 at 22:02
    
yup... Thanks a lot. –  Amihai Zivan Jan 8 '12 at 22:05

4 Answers 4

up vote 1 down vote accepted

$\mathbb{P}\left( {Y \leqslant y} \right) = \mathbb{P}\left( {\ln \frac{1} {{{{\left( {1 - X} \right)}^c}}} \leqslant y} \right) = \mathbb{P}\left( {X \leqslant 1 - {e^{ - \frac{y} {c}}}} \right) = \int_0^{1 - {e^{ - \frac{y} {c}}}} {\frac{1} {{1 - 0}}dx} = 1 - {e^{ - \frac{y} {c}}}$

Since CDF of exponentially distributed variable $Z$ with parameter $\lambda $ is given by $\mathbb{P}\left( {Z \leqslant z} \right) = 1 - {e^{ - \lambda z}}$, we conclude that $Y$ has exponential distribution with parameter $\frac{1}{c}$

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Thank you Alen. –  Amihai Zivan Jan 8 '12 at 22:09
1  
This is more complicated than what is really necessary. Sasha's answer, saying "Since $X$ is uniform, $\Pr(X\le x)=x$ for all $0\le x\le 1$" hits the nail on the head. –  Michael Hardy Jan 9 '12 at 1:24

You're almost there. Since $X$ is uniform, $\mathbb{P}(X \leqslant x) = x$ for all $0\leqslant x \leqslant 1$. Thus you established that $$ F_Y(y) = \mathbb{P}(Y \leqslant y) = \mathbb{P}\left(X \leqslant 1-\mathrm{e}^{-\frac{y}{c}} \right) = 1 - \exp\left(- \frac{y}{c} \right) $$ Now compare this with CDF of exponential distribution with parameter $\lambda$: $$ F_{\mathrm{Exp}(\lambda)}(y) = 1 - \exp(-\lambda y) $$ This proves that $Y \sim \mathrm{Exp}(c^{-1})$.

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indeed. Thanks. –  Amihai Zivan Jan 8 '12 at 22:07

Recall that the derivative of the distribution function of a continuous random variable $Y$ is the density of $Y$.

You've found that $F_Y(y)=1-e^{-y/c}$ for $y\ge0$.

Also $f_Y(y)=0$ for $y<0$ (since $ Y=\ln \bigl( \, (1-X)^c \bigr)\,)\ge 0 $ ). So $$ F_Y(y)= \cases { 1-e^{-y/c}&, y\ge0\cr0&,\text{otherwise} } $$

Differentiating $F_Y(y)$ gives $$ f(y) =\cases{{1\over c}e^{-y/c}&, y\ge 0\cr 0\vphantom{1\over c}&,\text{otherwise} . } $$

So $f_Y$ is the density of an exponential variable with parameter $1/c$.

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Here is a different approach. Given that $X\sim \mathcal{U}(0,1)$, $f_X(x)=1$.
From $y = \ln\left((1-x)^{-c}\right)$ we get $x=1-\mathrm{e}^{-\frac{y}{c}}$ and thus, $$\frac{\text{d}x}{\text{d}y}=\frac{1}{c}\mathrm{e}^{-\frac{y}{c}}~~~.$$ Thus $$\begin{align*} f_Y(y) & = f_X\left(1-\mathrm{e}^{-\frac{y}{c}}\right)\left|\frac{\text{d}x}{\text{d}y}\right|\\ & = \frac{1}{c}\mathrm{e}^{-\frac{y}{c}}~~, \end{align*} $$ which is what we wanted.

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