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

• November 10th 2011, 12:58 PM
wopashui
find the probability density function of the random variable Y=lnX
If X is an exponential random variable with parameter $\lambda$ $=1$, 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 $\lambda$=1, we have $f(x)=e^{-x}$, Y=lnX implies $X=e^y$, if I sub $e^y$ to f(x), it will be very hard to do the integral, any simple way to do this?
• November 11th 2011, 12:59 AM
mr fantastic
Re: find the probability density function of the random variable Y=lnX
Quote:

Originally Posted by wopashui
If X is an exponential random variable with parameter $\lambda$ $=1$, 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 $\lambda$=1, we have $f(x)=e^{-x}$, Y=lnX implies $X=e^y$, if I sub $e^y$ 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.
• November 11th 2011, 05:06 AM
CaptainBlack
Re: find the probability density function of the random variable Y=lnX
Quote:

Originally Posted by wopashui
If X is an exponential random variable with parameter $\lambda$ $=1$, 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 $\lambda$=1, we have $f(x)=e^{-x}$, Y=lnX implies $X=e^y$, if I sub $e^y$ 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:

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

CB