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Given the random variable X has a uniform distribution

$$f(x) = 1$$ $$\text{if }1<x<2$$

Find the probability distribution of the random variable: $$Y = -2\ln x$$

Could you anyone please help me in understanding how to solve the given problem, I don't know if I should be using PDF or CDF in the given problem, I have searched a lot for similar problems but I can't seem to find any and I got really confused with this one!

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  • $\begingroup$ You're given the PDF for $X$. Can you write down its CDF? If so, does that tell you anything about the CDF of $Y$? $\endgroup$ – Ravi Fernando Jun 2 '18 at 3:00
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    $\begingroup$ I have found the CDF for X and it is == to x-1, so I am not sure what to do with it. $\endgroup$ – Unix Jun 2 '18 at 3:04
  • $\begingroup$ Okay, so when $1 < x < 2$, the probability that $X < x$ is $x-1$. Can you use this to calculate the probability that $Y < y$ for some number $y$? $\endgroup$ – Ravi Fernando Jun 2 '18 at 3:09
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Lets consider the situation where we use the PDF. We know:

$$f_X(x)=1,\space y=-2\ln x \implies x=e^{-\frac y2}$$

Lets take the derivative with respect to $y$:

$$\frac {dx}{dy}=-\frac 12e^{-\frac y2}$$

Under the inverse transform method, we can use the fact that:

$$f_Y(y)=f_X(y) \left| \frac {dx}{dy} \right|$$ $$=f_X(-2 \ln x)\left|-\frac 12e^{-\frac y2} \right|$$ $$=1\cdot \frac 12e^{-\frac y2}$$ $$=\frac 12e^{-\frac y2}$$

The only part left is to find our bounds for the random variable $Y$.

From the equation $y=-2\ln x$:

$$x=1 \implies y=-2 \ln 1=0$$ $$x=2 \implies y=-2 \ln 2$$

We can check the density of $Y$ integrates to $1$:

$$\int_{-2 \ln 2}^0f_Y(y)dy=\int_{-2 \ln 2}^0\frac 12e^{-\frac y2}dy$$

$$=\left[-e^{- \frac y2}\right]_{-2 \ln 2}^0=-1-(-2)=1$$

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  • $\begingroup$ Could you elaborate more please, I want to see this example as solved so I can work on similar problems further. Thank you in advance! $\endgroup$ – Unix Jun 2 '18 at 3:09
  • $\begingroup$ Sure! I will go on to solve it. But if you want to check out some other examples, have a look at the link. It goes into a bit more detail than I will. math.arizona.edu/~jwatkins/f-transform.pdf $\endgroup$ – Tony Hellmuth Jun 2 '18 at 3:10

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