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I have a random variable U which is uniformly distributed over [0,1]; Now $$X=-2logU$$ Then what would be the P.D.F. of X? I know that P.D.F of U is 1 for [0,1] and 0 otherwise so the limits of X would be [0, $\infty]$ but how to determine its PDF(probability density function).

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Assuming that is the natural logarithm; $\;X=-2\log_\mathsf e(U)\;$ iif $\;U=\mathsf e^{-X/2}\;$.

Then by change of variables:

$$f_X(x) = f_U(\mathsf e^{-x/2})\left\lvert\dfrac{\operatorname{d} \mathsf e^{-x/2}}{\operatorname{d} x}\right\rvert$$


If you have not yet encountered this application of the chain rule we can show it the long way.

$$\begin{align} f_X(x) & =\frac{\operatorname d}{\operatorname dx} \mathsf P(X\leq x) \\[1ex] & = \frac{\operatorname d}{\operatorname dx} \mathsf P(-2\log_\mathsf e U\leq x) \\[1ex] & = \frac{\operatorname d}{\operatorname dx} \mathsf P(U\geq \mathsf e^{-x/2}) \end{align}$$

$$\begin{align} & = \frac{\operatorname d}{\operatorname dx} (1-F_U(\mathsf e^{-x/2})) \\[1ex] & = \frac{\operatorname d}{\operatorname dx} (1-\mathsf e^{-x/2})\;\mathbf 1_{\mathsf {exp}(-x/2)\in(0;1)} \\[1ex] & = -\frac{\operatorname d(\mathsf e^{-x/2})}{\operatorname dx}\;\mathbf 1_{x\in(0;\infty)} \\[1ex] & = \tfrac12\mathsf e^{-x/2}\;\mathbf 1_{x\in(0;\infty)} \end{align}$$


Thus $X\sim\mathcal{Exp}(1/2)$

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  • $\begingroup$ I have not used change of variables till now so is it that the function is inversely proportional to change of its argument (independent variable)? $\endgroup$ – Mayank Deora Dec 13 '15 at 5:03
  • $\begingroup$ It's an application of the chain rule. $F'_X(x) = F'_U(g(x))\; \lvert g'(x)\rvert$ when $U = g(X)$ and $g$ is an invertable function (over the support). $\endgroup$ – Graham Kemp Dec 13 '15 at 5:10
  • $\begingroup$ If you haven't encountered this yet, then probablyme's answer is excellent. @MayankChaturvedi $\endgroup$ – Graham Kemp Dec 13 '15 at 5:11
  • $\begingroup$ I understood the chain rule.It is applicable here because p.d.f is the differential of c.d.f. ;right? but why have you taken modulus of differentiation of $e^{-x/2}$? $\endgroup$ – Mayank Deora Dec 13 '15 at 7:30
  • $\begingroup$ A pdf is always non-negative. The modulus eliminated the 'direction' component of the CDF. @MayankChaturvedi $\endgroup$ – Graham Kemp Dec 14 '15 at 6:51
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  1. I would recommend using a change of variable on $$Y = -\log U.$$ People also like to use the cdf method $P(Y<y) = P(-\log U < y)$ to find the distribution. The distribution of this is

    $\text{Exp}(1)$

  2. Notice that now your problem has become $$X = 2 Y.$$ Using properties possibly shown in class or in the textbook, recognize that the distribution of $X$ is

    $\text{Exp}(1/2)$

Yes, you could have done a change of variable from the beginning or the cdf method on $P(Y< y) = P(-2\log U <y)$ from the beginning, but it's good to keep in mind everything you have learned up to now.

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All you need is that $\log$ is a strictly increasing transformation. Using that fact, we have: for $x\geq 0$, $$ \Pr(X\leq x)=\Pr(-2\log U\leq x)=\Pr(\log U\geq -x/2)=\Pr(U\geq e^{-x/2})=1-e^{-x/2}. $$ Thus, for $x\geq 0$, the pdf of $X$ at $x$ is $$ \frac{d}{dx}(1-e^{-x/2})=\frac{1}{2}e^{-x/2}. $$

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