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If $X$ and $Y$ are two independent random numbers from the interval $(0,1)$, then what is the PDF of $U = -2 \ln X$ and $V = -2\ln Y$?

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    $\begingroup$ If this is homework please add homework tag. $\endgroup$
    – mpiktas
    May 28, 2012 at 14:13
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    $\begingroup$ You haven't specified the densities for X and Y on [0,1]. The answer depends on what those distributions are. Did you intend for them to be uniform? $\endgroup$ May 28, 2012 at 15:00
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    $\begingroup$ @mp When the question is routine and artificial, as this one is, please feel free to add the homework tag. See the tag wiki for details. $\endgroup$
    – whuber
    May 28, 2012 at 19:09

2 Answers 2

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A general way for solving this type of problem is to find the CDF of the transformed variable in terms of the CDF of the original variable, then take the derivative to find the PDF of the transformed variable.

For example, assume $X$ is a random variable taking values in $(0,1)$ with PDF $f_X(x)$ and CDF $F_X(x)$. We would like to know the CDF of $U=-2\log(X)$.

$$P(U\leq u)=P(-2\log(X)\leq u)=P(X\geq e^{-\frac{u}{2}})=1-F_X(e^{-\frac{u}{2}})$$

Now we differentiate the CDF of $U$ with respect to $u$ and find that the PDF of $U$ is the following.

$$f_U(u)=\frac{d}{du}\left(1-F_X(e^{-\frac{u}{2}})\right)=-f_X\left(e^{-\frac{u}{2}}\right)\left(-\frac{1}{2}e^{-\frac{u}{2}}\right)=\frac{1}{2}e^{-\frac{u}{2}}f_X\left(e^{-\frac{u}{2}}\right)$$

Because $X$ takes on values between $0$ and $1$, $U$ takes on values between $0$ and $\infty$.

The same process can be completed for $V=-2\log(Y)$.

You can identify the distributions of $U$ and $V$, if they are well known, by inspection of the PDF or CDF.

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  • $\begingroup$ +1 Thanks Max. I took U=-ln(X) instead of -2ln(X) $\endgroup$ May 28, 2012 at 15:53
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If $X$ and $Y$ are independent uniform $U$ and $V$ are independent negative exponential with rate parameter $1/2$. You can solve this by change of variables.

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    $\begingroup$ Michael, not quite. $\endgroup$
    – cardinal
    May 28, 2012 at 15:10
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    $\begingroup$ It's actually an exponential with rate parameter $\frac{1}{2}$. $\endgroup$ May 28, 2012 at 15:12
  • $\begingroup$ Right guys. I forgot about the 2 in front of the log when I worked it out. I will change my answer. $\endgroup$ May 28, 2012 at 15:51

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