1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find cdf, pdf and expextation value of a random variable

  1. Nov 7, 2007 #1
    1. The problem statement, all variables and given/known data

    Let X represent the random choice of a real number on the interval [-1,1] which has a uniform distribution such that the probability density function is[tex]f_{X}(x)=\frac{1}{2}[/tex] when [tex]-1\leqx\leq1[/tex]. Let[tex] Y=X^{2}[/tex] a. Find the cumulative distribution [tex]F_{Y}(y)[/tex] b. the density function [tex]f_{Y}(y)[/tex] and c. the expected value [tex]E(Y)[/tex].

    2. Relevant equations

    my book gives a great explanation on how to change variables for joint distributions, but little is said about functions of one random variable, so I'm kind of at a loss here.

    3. The attempt at a solution

    first, if [tex]Y=X^{2}[/tex], then I want to say we need to find Y over the interval
    [0,1]. And integrating I have that:
    [tex]F_{X}(x)=\frac{x+1}{2}[/tex] which is [tex]P(X\leqx)[/tex]...
    now I want to say [tex]P(X\leqx)=P(\sqrt{Y}\leqx)[/tex]...

    i'm not sure where to go from here...
    can I just substitute [tex] \sqrt{y} [/tex] for x so I have:

    [tex]F_{Y}(y)=\frac{\sqrt{y}+1}{2}[/tex] ??
     
  2. jcsd
  3. Nov 8, 2007 #2

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    One way to look at the cumulative distribution function F(X) for some random process is F(x) = p(X<x). In other words, the value of the F(x) is the probability that a random number drawn from the distribution is smaller than x. If the random process has some lower bound and upper bound, the CDF F(x) must be identically zero below the lower bound and identically one above the upper bound. Your conjectured function does not exhibit this behavior. (Look at FY(0).)

    Using the above definition of the CDF, FY(y) = p(Y<y). Translate this expression to the original random variable and evaluate the resultant probability expression.
     
  4. Nov 8, 2007 #3
    part a:

    1. graph and draw a picture of y = x^2

    2. note that y is only valid from 0 to 1 ( since y = x^2 can only produce zero and positive numbers)

    3. to get the CDF formula, find the limits of the integral. y goes from 0 to x^2 and x goes from 0 to infinity. once the limits are found, plug the pdf into the CDF formula and then do the integration

    part b:

    part c:

    for part c there are two ways of doing this. use the expectation formula. or find the characteristic function and then evaluate -j times the first derivative of the characteristic function when the characteristic function's variable equals to zero.
     
    Last edited: Nov 8, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Find cdf, pdf and expextation value of a random variable
Loading...