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Math Help - Random variables - density functions

  1. #1
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    Question Random variables - density functions

    (a) If Z1 and Z2 are independent and have N(0,1) densities, let

    X1 = a11Z1 + a12Z2 + c1 and let X2 = a21Z1 + a22Z2 + c2

    where a11,a12,a21,a22,c1 and c2 are constants. Determine the joint density of X1 and X2.


    (b) Suppose that the random variable X on (0,1) has density

    f(x) = 3x2 0 < x < 1

    Determine the density functions of

    (i) U = X1/2
    (ii) V = -log(X)

    (c) Now suppose Y is uniformly distributed on (0,1) and independent of X. Determine the density of

    W = max(X,Y)
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  2. #2
    Member Last_Singularity's Avatar
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    Quote Originally Posted by GenericNameA View Post
    (a) If Z1 and Z2 are independent and have N(0,1) densities, let

    X1 = a11Z1 + a12Z2 + c1 and let X2 = a21Z1 + a22Z2 + c2

    where a11,a12,a21,a22,c1 and c2 are constants. Determine the joint density of X1 and X2.


    (b) Suppose that the random variable X on (0,1) has density

    f(x) = 3x2 0 < x < 1

    Determine the density functions of

    (i) U = X1/2
    (ii) V = -log(X)

    (c) Now suppose Y is uniformly distributed on (0,1) and independent of X. Determine the density of

    W = max(X,Y)
    Part B (i):
    We are given that f_X(x) = 3X^2, X \in [0,1]. And given that U = X^{1/2}, we need the distribution of U

    Step 1: Determine the domain of the random variable U. Since X is at most 1, U is at most 1. And similar, because X is at least 0, U is at least 0.
    So U \in [0,1] as well.

    Step 2: find the CDF of U: F_U(u)
    = P(U < u)

    = P(X^{1/2} < u)

    =P(X < u^2)

    =F_X(u^2)

    Step 3: find the pdf of U
    And since f_U(u) = \frac{\partial F_U}{\partial U} and we concluded that F_U(u) = F_X(u^2), it must be the case that:

    f_U(u) = \frac{\partial F_X}{\partial U}

    = f_x(u^2) \frac{\partial (u^2)}{\partial u}

    =2u f_x(u^2)

    =2u 3 (u^2)^2

    =6u^5

    Step 4: Vary that indeed this is a pdf:
    \int_0^1 6u^5 du = 1 as desired.

    Part B (ii):
    This one is a bit tougher. But we can proceed as before:

    Step 1: The boundaries of X are 0,1. So if we let V = - log(x), then the boundaries of V become 0 and positive infinity.
    So V \in [0, \infty]

    Step 2: find the CDF of V: F_V(v)
    = P(V < v)

    = P(-log(X) < v)

    = P(10^{-log(X)} < 10^v)

    = P(1/X < 10^v)

    = P(X < 10^{-v})

    = F_X(10^{-v})

    Step 3: find the PDF of V: f_V(v)
    = \frac{\partial F_X(10^{-v})}{\partial v}

    = -ln(10)(10^{-v}) f_x(10^{-v})

    =3 ln(10) (1/100)^v (10^{-v})

    Step 4: verify that this is indeed a valid pdf:
    \int_0^\infty 3 ln(10) (1/100)^v (10^{-v}) dv = 1 as desired.

    Part C:
    We proceed as before by finding the cdf of W:
    = P(W < w)

    Now, here's the trick. Since we have defined W = max(X,Y) or that W represents the larger of X and Y, it means that whenever W is less than a constant w, then it must be the case that both X and Y are less than that constant w. So we can determine that:
    = P(W < w)
    = P(X < w, Y < w)
    = P(X < w) P(Y < w) (by independent of X and Y)
    = F_X(w) F_Y(w)

    We know that F_X(x) = \int f_X(x) dx = \int 3X^2 dx= X^3 and that F_Y(y) = \int f_Y(y) dy = \int 1 dy = Y

    So F_X(w) F_Y(w) = (W^3)(W) = W^4

    We conclude that the cdf of W is: F_W(w) = W^4

    Then, of course, the pdf of W is: f_W(w) = 4W^3, W \in [0,1]

    Verify that this is a valid pdf and you are done.

    Part A:
    I am too tired lol...just scroll to the section that says "bivariate case" at: http://en.wikipedia.org/wiki/Bivariate_normal

    Plug in the appropriate variables into the formula...
    Last edited by Last_Singularity; January 14th 2009 at 07:25 PM.
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