# Expected Value of Ratio of Two Independent Random Variables

Discussion in 'Scientific Statistics Math' started by jp134711, Nov 28, 2010.

1. ### jp134711Guest

Hi,

I would like some verification to see if I have done this problem correctly:

Given two independent random variables, X and Y, both with exponential distribution with mean of 1, I am trying to find the expected value of Y/X.

Since X,Y are independent, E(Y/X) = E(Y)E(X^-1). I've used pretty much every algebraic/calculus manipulation I can think of (including straight defn of expected value, finding pdf of Y/X, Taylor series of X^-1, etc.) to solve this problem, and I keep getting E(Y/X) = Infinity.

I want to develop a gut feeling for this problem, but somehow I am unable to. Your help is much appreciated.

John

jp134711, Nov 28, 2010

2. ### jp134711Guest

As a follow-up, I also used Jensen's inequality, and this only showed me that E(Y/X) >= 1. Not much help either from this method.

jp134711, Nov 28, 2010

3. ### Jack TomskyGuest

As a follow-up, I also used Jensen's inequality, and
The easiest way to look at it is to note that an exponential with scale parameter one is the same as one-half times a chi-square with two degrees of freedom. Thus Y/X is an F with two and two degrees of freedom. The mean of F(2,2) is undefined (or actually infinity).

Jack
www.tomskystatistics.com

Jack Tomsky, Nov 28, 2010
4. ### Ray KoopmanGuest

Hint: An exponential variable is proportional to a chi-square variable
with 2 degrees of freedom. What do you know about the distribution of
the ratio of independent chi-square variables?

Ray Koopman, Nov 28, 2010
5. ### jp134711Guest

Thanks! That helps a lot in intuitively answering the problem.

jp134711, Nov 28, 2010
6. ### Luis A. AfonsoGuest

Luis A. Afonso, Nov 28, 2010
7. ### Herman RubinGuest

distribution with mean of 1, I am trying to find the expected value
of Y/X.
much every algebraic/calculus manipulation I can think of (including
straight defn of expected value, finding pdf of Y/X, Taylor series of
X^-1, etc.) to solve this problem, and I keep getting E(Y/X) = Infinity.
unable to. Your help is much appreciated.

ANY distribution with positive density at 0 does not
have a finite expected value for the reciprocal of
the random variable.

-- This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University. Herman
Rubin, Department of Statistics, Purdue University
Phone: (765)494-6054 FAX: (765)494-0558

Herman Rubin, Nov 30, 2010
8. ### Luis A. AfonsoGuest

The ratio of two Normal Standard N(0, 1) follows a Cauchy Distribution. . .

en.wikipedia.org/wiki/Normal_distribution

Luis

Luis A. Afonso, Nov 30, 2010
9. ### Luis A. AfonsoGuest

Date: May 8, 2008 2:28 AM
Author: Luis A. Afonso
Subject: How to simulate Cauchy samples

How to simulate Cauchy samples

********************************************
___Cauchy Distribution Function (From Wikipedia)

______F(x) = (1/pi) * arctan x + 1 / 2
______location (median) Density = 0
***********************************

_________x= tan [ pi* (F ? 0.5]__________(A)

(pi= 3.14159?)

Putting F = RND we are able to GENERATE samples of whichever size n having the Cauchy Density...

__________f(x) = 1 / [pi * ( 1 + x^2) ]

*********************

In order to find FRACTILES of the median:

__1___Using (A) synthesize the sample 1 sized n. Find its MEDIAN, w (1)

__2___Memorizing
Raking into account that (QBASIC) the indexed variables

___2a) Preparing the index:

____ i = Int (1000 * w (1)) + 4000

_____IF i <0 THEN i=0
_____IF i>8000 RHEN i=8000

___2b) Memorizing; _____W (i) = W (i) + 1.

:__3____After performing 400Â´000 times (for example) the calculations above indicted find the quantiles (for example 0.025 and 0.975).

RESULTS:

Limits of the 95% C.I. of Cauchy MEDIANS (as it was shown on Apr. 24, 2008 11:30 AM)

___n=5_______+/-2.012
_____7_________ 1.535
_____9_________ 1.274
____11_________ 1.110
____13_________ 0.989
____15_________ 0.905
____17_________ 0.838

(400Â´000 samples)

****************

Luis

Luis A. Afonso, Nov 30, 2010