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Probability density function of a random variable.

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data
    Let X be a posative random variable with probability density function f(x). Define the random variable Y by Y = X^2. What is the probability density function of Y? Also, find the density function of the random variable W = V^2 if V is a number chosen at random from the interval (-a,a) with a>0.


    2. Relevant equations



    3. The attempt at a solution
    I know the answer to the first part is f(sqrt(y))/2sqrt(y) and the answer to the second part is 1/(2a*sqrt(w)) but i have no idea how to get to these answers. I have looked through my book and notes many times and i'm still lost. Any help would be very much appreciated
     
  2. jcsd
  3. Apr 26, 2010 #2
    What is the probability density function? To solve think change of variables and integration.
     
  4. Apr 26, 2010 #3
    What do I integrate? The only way i could see how to do it was to derive but i still didn't know what to derive and how to plug it in.
     
  5. Apr 26, 2010 #4
    Can you define what a probability density function is in general terms?
     
  6. Apr 26, 2010 #5
    A probability density function of a continuous random variable is a function that describes the relative likelihood for this random variable to occur at a given point in the observation space.
     
  7. Apr 26, 2010 #6
  8. Apr 26, 2010 #7
    You do not use integration. I did Google search and the formula is in many places. Basically the probability contained in a differential area must be equal in each of the cases.

    [tex] |f(x)dx| = |f(y)dy| [/tex]
     
  9. Apr 26, 2010 #8
    ok but how does that get me to my answer?
     
  10. Apr 26, 2010 #9
  11. Apr 27, 2010 #10
    Ok thanks, how does it apply to the second part of the question then? I figured out the first part though.
     
  12. Apr 27, 2010 #11
    My course teaches us to derive from basic definitions. Not sure if its the best way but its the only way I know. :p

    (Cumulative) Distribution Function of Y
    = F(y)
    = P(Y < or = y)
    = P(X^2 < or = y)
    = P( -sqrt y < or = X < or = sqrt y)
    = [ 1 - P(x < -sqrt y) ] + P(x < or = sqrt y)
    = 1 - G(-sqrt Y) + G(sqrt Y)

    Where G is the Distribution Function of X.

    To get to the Density of Y, we need to differentiate F(y). Remember that G contains a function of Y so you need the chain rule.

    For W = V^2, V has a uniform distribution on (-a,a). Everything is the same except "Y" becomes "W" and "X" becomes "V". Use the density of V accordingly.

    P.S.
    your answer to the first part appears to be wrong. which implies the 2nd part is wrong too.
     
    Last edited: Apr 27, 2010
  13. Apr 27, 2010 #12
    Answers to the first and second part aren't wrong, those are the answers in the back of my book.
     
  14. Apr 27, 2010 #13
    Have you looked at my suggested method? After you differentiate the cumulative distribution function F(y) to get the probability density f(y), the answer is different from yours. Maybe there is a printing error. (happens all the time with textbooks)
     
  15. Apr 27, 2010 #14
    I did look at your method Legendre, and i think i just do the integral of f'(sqrt(y)), which by the chain rule gives me f(y)/(2sqrt(y)) which is the answer i need.
     
  16. Apr 27, 2010 #15
    You are right. The answer is correct. Sorry about the confusion!

    Cumulative Dist. = F(y) = 1 - G(-sqrt Y) + G(sqrt Y)

    Density = f(y) = d/dy [F(y)] = (1/2sqrt(y))[ g(-sqrt Y) + g(sqrt Y)]
    where g is the density of X.

    g(-sqrt Y) = 0 since X is a positive random variable.

    So, f(y) = (1/2sqrt(y))(g(sqrt Y)).


    P.S.
    I apologize for mixing up the notation! I know f is supposed to be the density of X in your question but for the sake of being consistent with my previous post, I let f be the density of Y and g be the density of X instead. Sorry!
     
  17. Apr 27, 2010 #16
    Cool thanks. Do you know how I would get the second part though. I know its got to be similar but in what step do i plug in the a. And why is it 1 over the answer rather than f(sqrt(W))? Thanks for the help
     
  18. Apr 27, 2010 #17
    You're welcomed man! Getting stuck at maths sucks! I got stuck for days recently and this forum helped me out, so I know how valuable maths help can be!

    W = V^2

    1) V has a continuous uniform density function over (-a,a) for a>0. Let g(v) be the density function for V. g(v) = 1/2a. The formula for density of continuous uniform distributions on interval (a,b) is "1/(b-a)".

    2) V is not a positive random variable because V is a number chosen from (-a,a). V < 0 for some outcomes.

    3) Considering the previous point, we need to re-derive the equation f(y) from our previous part. I am sticking to my notation for consistency. Apologies if its confusing.

    F(y) = 1 - G(-sqrtY) + G(sqrtY)

    Then f(y) = (1/2sqrt(y))[ g(-sqrt Y) + g(sqrt Y)

    4) We know g(v) = 1/2a for all v, so we plug it in and replace all y with w.

    f(y) = (1/2sqrt(w))(1/2a + 1/2a) = (1/2sqrt(w))(1/a) = 1/(2a*sqrt(w)) as required.
     
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