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

1. ## 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?

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 $\lambda$ $=1$, 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.

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 $\lambda$ $=1$, 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

### if x is an exponential random variable with parameter, compute the pdf of the random variable y defined by y=logx

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