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Suppose $X$ is a random variable that is uniformly distributed, and $f$ is a function such as $f:\mathbb{R}\rightarrow[-a, a]$ where $a$ is some constant.

How do I find the probability density function of $f(X)$? The function $f$ is not monotonic but can be assumed to be continuous and differentiable.

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  • $\begingroup$ Is $f$ monotonic? $\endgroup$ – Glen_b -Reinstate Monica Sep 25 '15 at 9:43
  • $\begingroup$ No, it would not be a monotonic function. $\endgroup$ – boston Sep 25 '15 at 9:46
  • $\begingroup$ Continuous? Piecewise continuous? Do you have a specific example in mind for $f$? $\endgroup$ – Glen_b -Reinstate Monica Sep 25 '15 at 9:48
  • $\begingroup$ Continuous and differentiable $\endgroup$ – boston Sep 25 '15 at 9:49
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Suppose that for each $z \in [-a.a]$, the equation $f(x) = z$ has a finite number of solutions in whatever interval $X$ happens to be uniformly distributed on, and that of these solutions, the derivative $\frac{\mathrm df(x)}{\mathrm dx}$ of $f(x)$ is nonzero only at $x_1, x_2, \ldots, x_n$. Note that $n$ might be different for different $z$'s and there might be no solutions at all in some cases. Then, for each $z$ in $[-a,a]$, $$f_Z(z) = \sum_{i=1}^n \frac{f_X(x_i)}{\left|\frac{\mathrm df(x)}{\mathrm dx}\right|_{x=x_i}} = \sum_{i=1}^n \frac{c}{\left|\frac{\mathrm df(x)}{\mathrm dx}\right|_{x=x_i}}\tag{1}$$ where $c$ is the common value of the pdf of $X$ at all points in the interval on which $X$ is uniformly distributed.

It might look like the right side of $(1)$ does not depend on $z$ at all, but it does so depend. Remember that the values of the $x_i$ (as well as their number!) depends on what $z$ you have chosen (that is, they are functions of $z$).

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You've given very general conditions, so I can only offer a very general answer:

If $X \sim P_X$ then we know that

$$P(f(X)\leq z) = P_X(X \in f^{-1}([0,z]))=\frac{Leb(f^{-1}([0,z]) \cap \mathbf{range}\; X)}{Leb(\mathbf{range}\; X)}$$

Where $Leb$ is the Lebesgue measure, $\mathbf{range}\; X$ is the range of $X$ and $x \in f^{-1}(z)$ is the (possibly non-compact) inverse-image of $f$ applied to the set $[0,z]$.

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Distribution is uniform. And the range for $x$ is $-a$ to $+a$. And the uniform distribution is defined as $f(x) = 1/(b-a)$ where $a<x<b$ i.e. upper limit is $b$ and lower limit is $a$. In your situation, lower limit is $-a$ and upper limit is $+a$ so the pdf will be $f(x)=1/(a-(-a)) = 1/(a+a) =1/(2a)$ where $x$ takes value from $-a$ to $+a$. And $f(x) = 0$ otherwise.

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  • $\begingroup$ Distribution is uniform. And the range for x is -a to +a. And the uniform distribution is defined as f(x) = 1/(b-a) where a<x<b i.e. upper limit is b and lower limit is a. In your situation, lower limit is -a and upper limit is +a so the pdf will be f(x)=1/(a-(-a)) = 1/(a+a) =1/(2a) where x takes value from -a to +a. And f(x) = 0 otherwise. $\endgroup$ – user89929 Sep 25 '15 at 10:06
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    $\begingroup$ This obviously false with such a modest restrictions on $f$. $\endgroup$ – mpiktas Sep 25 '15 at 12:09
  • $\begingroup$ Your answer was not being displayed correctly because it was not formatted appropriately; somehow, the HTM/whatever parser was choking on the math notation. Typesetting as MathJax corrected this. Please use MathJax formatting to avoid this problem in the future. A tutorial can be found here. meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Sycorax says Reinstate Monica Sep 25 '15 at 15:05

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