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

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    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?
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    Re: find the probability density function of the random variable Y=lnX

    Quote Originally Posted by wopashui View Post
    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.
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  3. #3
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    Re: find the probability density function of the random variable Y=lnX

    Quote Originally Posted by wopashui View Post
    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
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