Results 1 to 3 of 3

Math Help - Wondering if I did these correctly

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    1

    Wondering if I did these correctly

    Hello everyone! I solved (or rather, tried to) a couple of problems, and I was wondering if I attempted them correctly. Any feedback would be wonderful! Thank you in advance!

    1) Suppose the p.d.f. of X is

    f(x) = 1/2x for 0 < x < 2
    and 0 otherwise.

    Determine the p.d.f. of Y = 3X + 2.


    My solution:

    I found the inverse of Y = 3X + 2.
    This came out to be (y-2)/3 = x.

    Now, I substituted this value in for x in f(x).
    This gave f(x) = (3/2)/(y-2).

    Then, I found the derivative of (y-2)/3.
    This turned out to be 1/3.

    Multiplying (1/3) * (3/2)/(y-2) gave me (1/2)(y-2).

    So if the above work is correct, the p.d.f. of Y = 3X + 2 is

    (1/2)(y-2) for 2 < y < 8
    and 0 otherwise.

    2) Suppose the p.d.f. of X is

    f(x) = e^-x for x > 0,
    and 0 for x less than or equal to 0.

    Determine the p.d.f. of Y = X^(1/2)

    I found the inverse of Y = X^(1/2).
    This turned out to be Y^2 = X.
    Then, I subbed in y^2 for x in f(x).
    This gave e^(-y^2).
    Now, the derivative of y^2 is 2.
    Multiplying gives 2*e^(-y^2) as the p.d.f., where y lies between negative infinity and positive infinity.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Integrator020 View Post
    Hello everyone! I solved (or rather, tried to) a couple of problems, and I was wondering if I attempted them correctly. Any feedback would be wonderful! Thank you in advance!

    1) Suppose the p.d.f. of X is

    f(x) = 1/2x for 0 < x < 2
    and 0 otherwise.

    Determine the p.d.f. of Y = 3X + 2.

    My solution:

    I found the inverse of Y = 3X + 2.
    This came out to be (y-2)/3 = x.

    Now, I substituted this value in for x in f(x).
    This gave f(x) = (3/2)/(y-2).

    Then, I found the derivative of (y-2)/3.
    This turned out to be 1/3.

    Multiplying (1/3) * (3/2)/(y-2) gave me (1/2)(y-2).

    So if the above work is correct, the p.d.f. of Y = 3X + 2 is

    (1/2)(y-2) for 2 < y < 8
    and 0 otherwise.

    2) Suppose the p.d.f. of X is

    f(x) = e^-x for x > 0,
    and 0 for x less than or equal to 0.

    Determine the p.d.f. of Y = X^(1/2)

    I found the inverse of Y = X^(1/2).
    This turned out to be Y^2 = X.
    Then, I subbed in y^2 for x in f(x).
    This gave e^(-y^2).
    Now, the derivative of y^2 is 2.
    Multiplying gives 2*e^(-y^2) as the p.d.f., where y lies between negative infinity and positive infinity.
    1) Does \frac{1}{2} \int_2^8 (y-2) \, dy = 1 ....?

    2) Does 2 \int_{-\infty}^{+ \infty} e^{-y^2} \, dy = 1 ....?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Sep 2009
    Posts
    502
    Quote Originally Posted by Integrator020 View Post
    Hello everyone! I solved (or rather, tried to) a couple of problems, and I was wondering if I attempted them correctly. Any feedback would be wonderful! Thank you in advance!

    1) Suppose the p.d.f. of X is

    f(x) = 1/2x for 0 < x < 2
    and 0 otherwise.

    Determine the p.d.f. of Y = 3X + 2.


    My solution:

    I found the inverse of Y = 3X + 2.
    This came out to be (y-2)/3 = x.

    Now, I substituted this value in for x in f(x).
    This gave f(x) = (3/2)/(y-2).

    Then, I found the derivative of (y-2)/3.
    This turned out to be 1/3.

    Multiplying (1/3) * (3/2)/(y-2) gave me (1/2)(y-2).

    So if the above work is correct, the p.d.f. of Y = 3X + 2 is

    (1/2)(y-2) for 2 < y < 8
    and 0 otherwise.

    2) Suppose the p.d.f. of X is

    f(x) = e^-x for x > 0,
    and 0 for x less than or equal to 0.

    Determine the p.d.f. of Y = X^(1/2)

    I found the inverse of Y = X^(1/2).
    This turned out to be Y^2 = X.
    Then, I subbed in y^2 for x in f(x).
    This gave e^(-y^2).
    Now, the derivative of y^2 is 2.
    Multiplying gives 2*e^(-y^2) as the p.d.f., where y lies between negative infinity and positive infinity.
    If you can solve the 1st problem, you should be able to solve the 2nd one. The idea is the same.

    You are give f(x) as pdf, where x is a set of random numbers in [0,2] \in R.
    Y is another set of random number [2,8] \in R where Y(X).
    Your f(y) = f(Y(X=x))

    You have found you inverse function correctly, but after you substitute it into f(x) you made two mistakes: (1) you should get f(y); (2) you should have two terms in the function.

    The derivative that you got was meaning less because it does not say what slope you are looking for. Are you looking for the slope of f(y)? Where do you think the slope will take you? If you don't need it, why look for it?

    The total area under your f(x) and f(y), [a,b] \in R, both, must equal to 1, i.e, the integration of f(x) and f(y) [a,b] \in R must equal to 1, regardless, but first make sure that f(y) is correct before you go too far.
    Last edited by novice; October 4th 2009 at 08:19 AM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Wondering if I did this right..
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 13th 2008, 08:58 PM
  2. Wondering
    Posted in the Business Math Forum
    Replies: 3
    Last Post: September 1st 2008, 06:58 AM
  3. Hi, new here! Was wondering if I am right, I would
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: June 26th 2008, 07:02 PM
  4. just wondering
    Posted in the Statistics Forum
    Replies: 4
    Last Post: November 12th 2007, 09:55 AM

Search Tags


/mathhelpforum @mathhelpforum