# Thread: find the probability density function of the random variable Y=lnX

If X is an exponential random variable with parameter $\displaystyle \lambda$$\displaystyle =1, find the probability density function of the random variable Y=lnX. HINT: Find the cummulative distribution function. I don't understand what the hint saying, find the cdf of what? when \displaystyle \lambda=1, we have \displaystyle f(x)=e^{-x}, Y=lnX implies \displaystyle X=e^y, if I sub \displaystyle e^y to f(x), it will be very hard to do the integral, any simple way to do this? 2. ## Re: find the probability density function of the random variable Y=lnX Originally Posted by wopashui If X is an exponential random variable with parameter \displaystyle \lambda$$\displaystyle =1$, find the probability density function of the random variable Y=lnX.
HINT: Find the cummulative distribution function.

I don't understand what the hint saying, find the cdf of what? when $\displaystyle \lambda$=1, we have $\displaystyle f(x)=e^{-x}$, Y=lnX implies $\displaystyle X=e^y$, if I sub $\displaystyle e^y$ to f(x), it will be very hard to do the integral, any simple way to do this?
Well, here is a hint for the hint: Research the probability integral transform theorem.

3. ## Re: find the probability density function of the random variable Y=lnX

Originally Posted by wopashui
If X is an exponential random variable with parameter $\displaystyle \lambda$$\displaystyle =1$, find the probability density function of the random variable Y=lnX.
HINT: Find the cummulative distribution function.

I don't understand what the hint saying, find the cdf of what? when $\displaystyle \lambda$=1, we have $\displaystyle f(x)=e^{-x}$, Y=lnX implies $\displaystyle X=e^y$, if I sub $\displaystyle e^y$ to f(x), it will be very hard to do the integral, any simple way to do this?
The hint is suggesting you to find:

$\displaystyle F_Y(y)=P(Y\le y)=P(\ln(X)\le y)=P(X\le e^y)$

CB