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I have a question. I'm given a continuous random variable X with the pdf:

$$f_X(x) = {1\over(1+\pi^2x^2)}\cdot I_{(-\infty, 0)\bigcup(0, \infty)}(x) $$

and asked to find the pdf and cdf of $Y = {1\over X}$. I'm given the hint that $F_Y(-\pi)=1/4$.

I'm having trouble solving for the cdf of the r.v. $Y$.

I get that $F_Y(-\pi)={3\over 4}$ (if $\arctan(-1) = -{\pi\over 4}$) or $-{1\over 4}$ (if $\arctan(-1) = {3\pi\over 4}$), however. This intuitively doesn't make much sense to me, even though I'm not seeing what I'm doing wrong with the integration.

Any ideas? Thanks!

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  • $\begingroup$ In your integral for $F_Y(y)$, why are your limits $\int_{1/y}^\infty$? Seems to me that they should be $\int_{-\infty}^{1/y}$. $\endgroup$ – alexwlchan Mar 9 '14 at 21:37
  • $\begingroup$ Because of the $P({1\over y} \le X)$ statement. But I think I have it. Per Andre's suggestion below, you (presumably) flip the inequality, which makes the limits what you suggest (and what works). $\endgroup$ – Guest Mar 9 '14 at 21:44
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Hint: We will need to take some care when $y$ is negative, since division then reverses inequalities.

But let's take $y$ positive. What is the probability that $Y\le y$?
This event happens if either $X\lt 0$ or $X\ge 1/y$.

By symmetry $\Pr(X\lt 0)=\frac{1}{2}$. Your integral expression for $\Pr(X\ge 1/y)$ is correct.

Added: For negative $y$, I would suggest not doing a separate calculation, negative numbers are treacherous. If $y$ is negative, then by symmetry the probability that $Y$ is $\le y$ is the same as the probability that $Y\ge |y|$.

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  • $\begingroup$ Thanks! That's what I figure (re: your hint), but not sure where to run with that. Definitely get the symmetry part -- I figure that's why the integral produces the ${1\over 2}$ term. Just not sure about the rest. (If the arctan() term was positive instead of negative, it'd work out. But that's voodoo math to me unless I can actually compute that.) $\endgroup$ – Guest Mar 9 '14 at 21:39
  • $\begingroup$ Are you hinting that I should add ${1\over 2}$ to my current expression? That'd work for $arctan(-1)= {3\pi\over 4}$. $\endgroup$ – Guest Mar 9 '14 at 21:49
  • $\begingroup$ For positive $y$, we have $F_Y(y)=\frac{1}{2}+\int_{1/y}^\infty \frac{dx}{1+\pi^2x^2}$. You integrated correctly, we get $F_Y(y)=1-\frac{1}{\pi}\arctan(\pi/y)$. This makes sense, it approaches $1$ as $y$ goes to infinity. $\endgroup$ – André Nicolas Mar 9 '14 at 21:51
  • $\begingroup$ It is not quite a hint, I think it says explicitly you should add $\frac{1}{2}$. $\endgroup$ – André Nicolas Mar 9 '14 at 21:52
  • $\begingroup$ Thank you! I get your explanation conceptually, just didn't know how I could reconcile that with the math. But this all makes sense. $\endgroup$ – Guest Mar 9 '14 at 21:57

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