Cauchy Distribution

thomas_donald · #1 $Old$ October 28th, 2009, 10:35 AM

Can I get help on the following HW question:
If X is a standard Cauchy Random Variable then 1/X is also a standard Cauchy Random Variable. How can it be proved?
Is it correct to say that: P(1/X<x) = P(X>1/x) = 1 - P(X<1/x) for x>0?
What about if x<0

theodds · #2 $Old$ October 28th, 2009, 11:18 AM

You seem to be on the right track. I would start with defining $Y = \frac{1}{X}$ , and begin with $F_{Y}(y) = P \left (Y \le y \right ) = P\left (\frac{1}{X} \le y \right ) = P \left (X \ge \frac{1}{y} \right ) = 1 - F_{X}(1 / y)$ , then take the derivative and see if you recognize it.

thomas_donald · #3 $Old$ October 28th, 2009, 12:04 PM

Thanks a lot. But what I am not clear about is that Cauchy Dist is from -infinity to +infinity. So is it correct that

both when y is positive and when y is negative. I would think that if y<0
then I should have P(Y<y) = P(1/X<y) = P(X<1/y) but then the derivative will
be negative???

matheagle · #4 $Old$ October 28th, 2009, 12:13 PM

Use $f_Y(y)=f_X(x)\biggl|{dx\over dy}\biggr|$

$={1\over \pi(1+{1\over y^2})}\biggl|{-1\over y^2}\biggr|$

$={1\over \pi(y^2+1)}$

theodds · #5 $Old$ October 28th, 2009, 01:26 PM

Quote:

Originally Posted by matheagle $View Post$

Use $f_Y(y)=f_X(x)\biggl|{dx\over dy}\biggr|$

$={1\over \pi(1+{1\over y^2})}\biggl|{-1\over y^2}\biggr|$

$={1\over \pi(y^2+1)}$

I'm not sure this works, since g(X) = 1/X is not monotone over the whole real line, and moreover is undefined for part of the support of X. The method DOES give the right answer, but I think if you are going to use it, you should handle to problem point X = 0. I think it's easier just to use work directly with the cdf.

theodds · #6 $Old$ October 28th, 2009, 01:31 PM

Quote:

Originally Posted by thomas_donald $View Post$

Thanks a lot. But what I am not clear about is that Cauchy Dist is from -infinity to +infinity. So is it correct that

both when y is positive and when y is negative. I would think that if y<0
then I should have P(Y<y) = P(1/X<y) = P(X<1/y) but then the derivative will
be negative???

Taking the reciprocal reverses the ordering regardless of where you are on the real line; e.g. -4 < -3 and -1/4 > -1/3.

thomas_donald · #7 $Old$ October 28th, 2009, 02:10 PM

Thank you very much. Now I get it.

matheagle · #8 $Old$ October 29th, 2009, 12:21 AM

Quote:

Originally Posted by theodds $View Post$

I'm not sure this works, since g(X) = 1/X is not monotone over the whole real line, and moreover is undefined for part of the support of X. The method DOES give the right answer, but I think if you are going to use it, you should handle to problem point X = 0. I think it's easier just to use work directly with the cdf.

I've never worried about a set of measure zero
and 1/x is decreasing for both x<0 and x>o, so that part is fine
This technique is just the derivative of the technique using F(x).
If you worry about it, just do one case at a time, x<0 and then x>0.

thomas_donald · #9 $Old$ October 29th, 2009, 08:45 AM

Thank you very much for the help.