Let fYt Y Y 6 eX Compute the probability density function

Let fY(t)= Y : Y = 6 e^X . Compute the probability density function of Y be the random variable defined by lambda = 6 , and let X be an exponential random variable with parameter

Solution

f(x) = 6e^-x, x>=0 is distribution of x

y = 6e^x

Or x = ln (y/6)

Hence f(y) = 6e^-lny/6

  ^{= k36/y ,}

^{y varies from 1 to e}

^{As total prob =1}

^{we have k = 1/36}

^{and pdf of y = 1/y , y from 1 to e}