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I want to find the CDF of a a random variable $X(\omega) = e^\omega$, with the sample space $\Omega = [-1,1]$. The outcomes of $\Omega$ are uniformly distributed.

What I've managed so far is to get to this point:

$F_X(x) = P(X<= x) = P(e^w <= x) = P(w <= ln(x))$

But I don't know where to go from here. I can't find anything about it in my books or from what I could gather on the net. All the help is welcome!

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First figure out what values $X$ can take on. – Dilip Sarwate Oct 16 '13 at 23:24
up vote 0 down vote accepted

Let $W$ be uniformly distributed on $[-1,1]$. Let $X=e^W$. We want the cdf of $X$. We have $$F_X(x)=\Pr(X\le x)=\Pr(e^W\le x)=\Pr(W\le \ln x).$$ So far, apart from a minor notational change, this is exactly what you wrote.

If $\ln x\lt -1$, then $\Pr(W\le \ln x)=0$. Thus $F_X(x)=0$ if $x\lt e^{-1}$.

If $\ln x\ge 1$, then $\Pr(W\le ln x)=1$. Thus $F_X(x)=1$ if $x\ge e$.

For $-1\le \ln x\lt 1$, we have $\Pr(W\le \ln x)=\frac{\ln x-(-1)}{2}$.

Thus in the interval $[e^{-1},e)$ we have $F_X(x)=\frac{\ln x+1}{2}$.

If the density function $f_X(x)$ is desired, differentiate.

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