Expected Value of Ratio of Two Independent Random Variables

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

  1. jp134711

    jp134711 Guest

    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
    #1
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  2. jp134711

    jp134711 Guest

    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
    #2
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  3. jp134711

    Jack Tomsky Guest

    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
    #3
  4. jp134711

    Ray Koopman Guest

    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
    #4
  5. jp134711

    jp134711 Guest

    Thanks! That helps a lot in intuitively answering the problem.
     
    jp134711, Nov 28, 2010
    #5
  6. Luis A. Afonso, Nov 28, 2010
    #6
  7. jp134711

    Herman Rubin Guest

    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
    #7
  8. 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
    #8
  9. 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
    ***********************************
    This leads to

    _________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 400´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

    (400´000 samples)

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

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