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  • Transformation question

    I would appreciate if anyone could help with this question. It is from Probability and Statistics with Applications: A Problem Solving Text Asimow and Mawell
    8-3
    Suppose that the random variable X has an exponential distribution with mean one. Let the random variable Y= sqrt (X)

    (a) Find the probability density function for Y.
    (b) Find the expected value for Y.

    For part (a) I got the density function for Y: f(y)= (density function for X)*(dx/dy) = e^-x * 2y= 2ye^(-y^2)

    Using that answer I attempted to solve part (b) by doing the integral from 0 to infinity of y*f(y). But I cannot figure out a solution using integration tables or integration by parts. I suspect I've made an error in the density function of Y.

  • #2
    Originally posted by bhopkins View Post
    I would appreciate if anyone could help with this question. It is from Probability and Statistics with Applications: A Problem Solving Text Asimow and Mawell
    8-3
    Suppose that the random variable X has an exponential distribution with mean one. Let the random variable Y= sqrt (X)

    (a) Find the probability density function for Y.
    (b) Find the expected value for Y.

    For part (a) I got the density function for Y: f(y)= (density function for X)*(dx/dy) = e^-x * 2y= 2ye^(-y^2)

    Using that answer I attempted to solve part (b) by doing the integral from 0 to infinity of y*f(y). But I cannot figure out a solution using integration tables or integration by parts. I suspect I've made an error in the density function of Y.
    Is the answer \sqrt{\pi}/4? If so you might have to get it in form of the standard normal and then see which factor is missing and use the symmetry of standard normal random variable.

    Comment


    • #3
      Thank you for replying. I finally figured it out sorry I didn't update the thread.

      the answer is sqrt(pi)/2 I had to use an improper integral that converges I found in an earlier chapter it was the integral from negative infinity to positive infinity or e^(-(z/2)^2) is equal to sqrt(2 pi). Or half of that since the integral starts at zero. I integrated sqrt(x)*f(x) instead of the transformation y*f(y) and did a u-substitution. . . then integration by parts.

      As it turns out after reading more in that previous chapter, there is a property of gamma distribution, The integral from zero to infinity or x^(a-1)*e^(-(1/b)x) is equal to b^(a) * gamma(a) . which would have be a lot quicker.
      There was another integral I came across that would have helped me if I tried to integrate y*f(y) but I can't find it. I will go back and look at your way just to see, but the key was finding a improper integral with e in it that equaled some thing with pi in it. I was looking at a calculus book for the longest but i guess it was a probability specific problem.

      I very much appreciate your help. It seems you have no more exams, so I'll try to pass the kindness on by help someone else if I am able.

      Comment


      • #4
        Originally posted by bhopkins View Post
        Thank you for replying. I finally figured it out sorry I didn't update the thread.

        the answer is sqrt(pi)/2 I had to use an improper integral that converges I found in an earlier chapter it was the integral from negative infinity to positive infinity or e^(-(z/2)^2) is equal to sqrt(2 pi). Or half of that since the integral starts at zero. I integrated sqrt(x)*f(x) instead of the transformation y*f(y) and did a u-substitution. . . then integration by parts.

        As it turns out after reading more in that previous chapter, there is a property of gamma distribution, The integral from zero to infinity or x^(a-1)*e^(-(1/b)x) is equal to b^(a) * gamma(a) . which would have be a lot quicker.
        There was another integral I came across that would have helped me if I tried to integrate y*f(y) but I can't find it. I will go back and look at your way just to see, but the key was finding a improper integral with e in it that equaled some thing with pi in it. I was looking at a calculus book for the longest but i guess it was a probability specific problem.

        I very much appreciate your help. It seems you have no more exams, so I'll try to pass the kindness on by help someone else if I am able.
        Using that answer I attempted to solve part (b) by doing the integral from 0 to infinity of y*f(y). But I cannot figure out a solution using integration tables or integration by parts sizzling hotslot
. I suspect I've made an error in the density function of Y.

        Comment

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