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Math Help - Just 2 questions...

  1. #1
    Bar0n janvdl's Avatar
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    Just 2 questions...

    Hello everyone
    Would appreciate help with these two, especially the second one.

    The joint densities of S = X + Y and T = \frac{X}{Y} is given by:

    f_{ST}(s,t) = \frac{\lambda ^2 s}{(1+t)^2} e^{- \lambda s}

    s \geq 0 \ ; \ t \geq 0

    where X and Y are independent exponential random variables with parameter \lambda

    Prove that S and T are independent by finding f_{S}(s) and f_{T}(t) and showing that (f_{S}(s))(f_{T}(t)) = f_{ST}(s,t)
    Let U_{1} and U_{2} be independent and uniform on [0,1]. Find and sketch the density function of S = U_{1} + U_{2}
    Once again I'd also like to ask if someone can suggest a book with sufficient examples and explanations regarding this kind of work. The book I have has very few examples, and even just as few explanations.
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  2. #2
    Grand Panjandrum
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    For the second the density of S is the convolution of the densities of U_1 and U_2, which gives the triangular distribution on [0,2] (symmetric about 1)

    RonL
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  3. #3
    Grand Panjandrum
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    The joint densities of S = X + Y and T = \frac{X}{Y} is given by:
    f_{ST}(s,t) = \frac{\lambda ^2 s}{(1+t)^2} e^{- \lambda s}
    s \geq 0 \ ; \ t \geq 0
    where X and Y are independent exponential random variables with parameter \lambda


    The marginals are:

    f_S(s)=\int_0^{\infty} f_{ST}(s,t) dt=s \lambda^2 e^{-\lambda s}

    and:

    f_T(t)=\int_0^{\infty} f_{ST}(s,t) ds=\frac{1}{(t+1)^2}

    and the rest follows (this is obvious as f_{ST} seperates into a product of a function of s and a function of t)

    RonL
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  4. #4
    Senior Member tukeywilliams's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    The joint densities of S = X + Y and T = \frac{X}{Y} is given by:
    f_{ST}(s,t) = \frac{\lambda ^2 s}{(1+t)^2} e^{- \lambda s}
    s \geq 0 \ ; \ t \geq 0
    where X and Y are independent exponential random variables with parameter \lambda


    The marginals are:

    f_S(s)=\int_0^{\infty} f_{ST}(s,t) dt=s \lambda^2 e^{-\lambda s}

    and:

    f_T(t)=\int_0^{\infty} f_{ST}(s,t) ds=\frac{1}{(t+1)^2}

    and the rest follows (this is obvious as f_{ST} seperates into a product of a function of s and a function of t)

    RonL
    Shouldn't they both be negative?
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  5. #5
    Bar0n janvdl's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    For the second the density of S is the convolution of the densities of U_1 and U_2, which gives the triangular distribution on [0,2] (symmetric about 1)

    RonL
    Still a little bit lost on this one. The lecturer said we'll hear about convolutions a bit later, and not to worry about it yet... (I'll have to go look up what a triangular distribution is as well.)

    Quote Originally Posted by CaptainBlack View Post
    The joint densities of S = X + Y and T = \frac{X}{Y} is given by:
    f_{ST}(s,t) = \frac{\lambda ^2 s}{(1+t)^2} e^{- \lambda s}
    s \geq 0 \ ; \ t \geq 0
    where X and Y are independent exponential random variables with parameter \lambda


    The marginals are:

    f_S(s)=\int_0^{\infty} f_{ST}(s,t) dt=s \lambda^2 e^{-\lambda s}

    and:

    f_T(t)=\int_0^{\infty} f_{ST}(s,t) ds=\frac{1}{(t+1)^2}

    and the rest follows (this is obvious as f_{ST} seperates into a product of a function of s and a function of t)

    RonL
    This has helped a lot, thank you Captain.

    Quote Originally Posted by tukeywilliams View Post
    Shouldn't they both be negative?
    Here's just a guess... they're exponential so they must be positive?

    EDIT: Wait, for some bizarre reason I also think they must be both negative. I tried working it out by myself just now...
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by tukeywilliams View Post
    Shouldn't they both be negative?
    They are pdf's of the marginal distributions and so have to be positive (and integrate up to 1).

    RonL
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by janvdl View Post
    Still a little bit lost on this one. The lecturer said we'll hear about convolutions a bit later, and not to worry about it yet... (I'll have to go look up what a triangular distribution is as well.)
    ..
    This is a bit fiddly to do as the convolution is peicewise continuous, but the convolution is:

    f_S(s)=\int_{-\infty}^{\infty}p_{U_1}(u_1)p_{U_2}(s-u_1) \ dx_1

    and using the properties of the uniform density:

    f_S(s)=\int_{-\infty}^{\infty}p_{U_1}(u_1)p_{U_2}(s-u_1) \ dx_1=\int_{0}^{1}p_{U}(s-u) \ du

    RonL
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  8. #8
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    Quote Originally Posted by CaptainBlack View Post
    The joint densities of S = X + Y and T = \frac{X}{Y} is given by:
    f_{ST}(s,t) = \frac{\lambda ^2 s}{(1+t)^2} e^{- \lambda s}
    s \geq 0 \ ; \ t \geq 0
    where X and Y are independent exponential random variables with parameter \lambda


    The marginals are:

    f_S(s)=\int_0^{\infty} f_{ST}(s,t) dt=s \lambda^2 e^{-\lambda s}

    and:

    f_T(t)=\int_0^{\infty} f_{ST}(s,t) ds=\frac{1}{(t+1)^2}

    and the rest follows (this is obvious as f_{ST} seperates into a product of a function of s and a function of t)

    RonL
    To Jan:

    A reference you might find helpful:

    http://www.dartmouth.edu/~chance/tea...k/Chapter7.pdf


    Given your current knowledge, I think it’s instructive for you to think about how to calculate the pdf of T = \frac{X}{Y}.

    The cdf of T is given by F (t) = \Pr(T \leq t) = \Pr \left( \frac{X}{Y} \leq t \right) = \Pr(X \leq t Y) since Y > 0.

    Therefore:

    F (t) = \int_{y = 0}^{\infty} \int_{x=0}^{x=ty} g(x,y) \, dx \, dy

    where g(x, y) is the joint pdf of X and Y.

    Since X and Y are independent random variables: g(x,y) = \frac{1}{\lambda^2} e^{-x/ \lambda} \, e^{-y/ \lambda} for x > 0, y > 0, and is zero elsewhere.

    Therefore:


    F (t) = \frac{1}{\lambda^2} \int_{y = 0}^{+ \infty} e^{-y/ \lambda} \int_{x=0}^{x=ty} e^{-x/ \lambda} \, dx \, dy


     = \frac{1}{\lambda} \int_{y = 0}^{+ \infty} e^{-y/ \lambda} \left[ - e^{-x/ \lambda} \right]_{0}^{ty} \, dy


     = \frac{1}{\lambda} \int_{y = 0}^{+ \infty} e^{-y/ \lambda} \left( 1 - e^{-ty/ \lambda} \right) \, dy


     = \frac{1}{\lambda} \int_{y = 0}^{+ \infty} e^{-y/ \lambda} - e^{-y(1+t)/ \lambda} \, dy


     = \left[ -e^{-y/ \lambda} + \frac{1}{1+t} \, e^{-y(1+t)/ \lambda} \right]_{0}^{+ \infty}


    = 1 - \frac{1}{1 + t}.


    Therefore the pdf of T is given by f(t) = \frac{dF}{dt} = \frac{1}{(1 + t)^2}.


    This is (of course) in agreement with the marginal distribution found using the given joint pdf of S and T:

    f(t) = \int_{0}^{+ \infty} \frac{\lambda^2 s}{(1 + t)^2}\, e^{- \lambda s} \, ds = \frac{\lambda^2}{(1 + t)^2} \, \int_{0}^{+ \infty}  s \, e^{- \lambda s} \, ds .....

    After studying the definition of convolution and looking at CaptainB’s example you should attempt to get the pdf of S = X + Y.
    Last edited by mr fantastic; August 10th 2008 at 10:38 PM.
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  9. #9
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    Quote Originally Posted by janvdl View Post
    Hello everyone
    Would appreciate help with these two, especially the second one.


    Quote:
    Let and be independent and uniform on . Find and sketch the density function of

    Once again I'd also like to ask if someone can suggest a book with sufficient examples and explanations regarding this kind of work. The book I have has very few examples, and even just as few explanations.
    The second one is done as Example 7.3 in the link I provided. It confirms the answer given by CaptainB (not that confirmation was required lol!)
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    Quote Originally Posted by CaptainBlack View Post
    For the second the density of S is the convolution of the densities of U_1 and U_2, which gives the triangular distribution on [0,2] (symmetric about 1)

    RonL
    To Jan: It can also be done without ever knowing that you're essentially doing a convolution. To simplify the notation, I'll use S = U + V.

    Since U and V are independent, their joint pdf is g(u, v) = (1)(1) = 1 for 0 \leq u \leq 1, 0 \leq v \leq 1, and zero elsewhere.

    The cdf of S is given by F(s) = \Pr(S \leq s) = \Pr(U + V \leq s) = \Pr(U \leq -V + s).

    There are two cases to consider: 0 \leq s \leq 1 and 1 \leq s \leq 2.


    Case 1: 0 \leq s \leq 1.

    F(s) = \int_{v=0}^{s} \int_{u=0}^{-v+s} du \, dv = \int_{0}^{s} [u]_{0}^{-v+s} \, dv = \int_{0}^{s} -v + s \, dv = \left[ -\frac{1}{2} v^2 + sv\right]_{0}^{s} = \frac{1}{2} s^2.


    Therefore f(s) = \frac{dF}{ds} = s for 0 \leq s \leq 1.



    Case 2: 1 \leq s \leq 2.

    F(s) = \int_{v=0}^{s-1} \int_{u=0}^{1} du \, dv + \int_{v=s-1}^{1} \int_{u=0}^{u=-v+s} du \, dv = \, .....


    I'll leave the mopping up to you. You should get f(s) = \frac{dF}{ds} = 2 - s for 1 \leq s \leq 2.


    And of course f(s) = 0 elsewhere. These are the same results as CaptainB got using the convolution theorem.
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  11. #11
    Bar0n janvdl's Avatar
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    Quote Originally Posted by mr fantastic View Post
    To Jan: It can also be done without ever knowing that you're essentially doing a convolution. To simplify the notation, I'll use S = U + V.

    Since U and V are independent, their joint pdf is g(u, v) = (1)(1) = 1 for 0 \leq u \leq 1, 0 \leq v \leq 1, and zero elsewhere.

    The cdf of S is given by F(s) = \Pr(S \leq s) = \Pr(U + V \leq s) = \Pr(U \leq -V + s).

    There are two cases to consider: 0 \leq s \leq 1 and 1 \leq s \leq 2.


    Case 1: 0 \leq s \leq 1.

    F(s) = \int_{v=0}^{s} \int_{u=0}^{-v+s} du \, dv = \int_{0}^{s} [u]_{0}^{-v+s} \, dv = \int_{0}^{s} -v + s \, dv = \left[ -\frac{1}{2} v^2 + sv\right]_{0}^{s} = \frac{1}{2} s^2.


    Therefore f(s) = \frac{dF}{ds} = s for 0 \leq s \leq 1.



    Case 2: 1 \leq s \leq 2.

    F(s) = \int_{v=0}^{s-1} \int_{u=0}^{1} du \, dv + \int_{v=s-1}^{1} \int_{u=0}^{u=-v+s} du \, dv = \, .....


    I'll leave the mopping up to you. You should get f(s) = \frac{dF}{ds} = 2 - s for 1 \leq s \leq 2.


    And of course f(s) = 0 elsewhere. These are the same results as CaptainB got using the convolution theorem.
    Thank you Mr F. I left this one out on the paper though. But I was the only one in class who did work.
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    Quote Originally Posted by mr fantastic View Post
    To Jan:

    A reference you might find helpful:

    http://www.dartmouth.edu/~chance/tea...k/Chapter7.pdf


    Given your current knowledge, I think it’s instructive for you to think about how to calculate the pdf of T = \frac{X}{Y}.

    The cdf of T is given by F (t) = \Pr(T \leq t) = \Pr \left( \frac{X}{Y} \leq t \right) = \Pr(X \leq t Y) since Y > 0.

    Therefore:

    F (t) = \int_{y = 0}^{\infty} \int_{x=0}^{x=ty} g(x,y) \, dx \, dy

    where g(x, y) is the joint pdf of X and Y.

    Since X and Y are independent random variables: g(x,y) = \frac{1}{\lambda^2} e^{-x/ \lambda} \, e^{-y/ \lambda} for x > 0, y > 0, and is zero elsewhere.
    May I know how does one get the above g(x,y)?
    I got g(x,y) = {y^2 {\lambda}^2 \over (x + y)} e^{-x/ \lambda} \, e^{-y/ \lambda} after subst for s and t.
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    Quote Originally Posted by chopet View Post
    May I know how does one get the above g(x,y)?
    I got g(x,y) = {y^2 {\lambda}^2 \over (x + y)} e^{-x/ \lambda} \, e^{-y/ \lambda} after subst for s and t.
    ...... and are independent exponential random variables with parameter ......
    Therefore g(x, y) = f(x) f(y) = ......
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    Ok...think I missed a Jacobian term when converting from (s,t) -> (x,y)

     J = {{\partial s} \over {\partial x}} {{\partial t} \over {\partial y}} - {{\partial s} \over {\partial y}} {{\partial t} \over {\partial x}} = {{x+y} \over {y^2}}

    <br />
g(x,y) = {y^2 {\lambda}^2 \over (x + y)} e^{-x/ \lambda} \, e^{-y/ \lambda}*J =  {\lambda}^2 e^{-x/ \lambda} \, e^{-y/ \lambda}

    still, the  {\lambda}^2 term is wrong...
    Last edited by chopet; September 1st 2008 at 06:58 PM.
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    Quote Originally Posted by chopet View Post
    Ok...think I missed a Jacobian term when converting from (s,t) -> (x,y)

     J = {{\partial s} \over {\partial x}} {{\partial t} \over {\partial y}} - {{\partial s} \over {\partial y}} {{\partial t} \over {\partial x}} = {{x+y} \over {y^2}}

    <br />
g(x,y) = {y^2 {\lambda}^2 \over (x + y)} e^{-x/ \lambda} \, e^{-y/ \lambda}*J = {\lambda}^2 e^{-x/ \lambda} \, e^{-y/ \lambda}

    still, the  {\lambda}^2 term is wrong...
    ...... and are independent exponential random variables with parameter ......
    Therefore g(x, y) = f(x) f(y) = ......
    where f(x) = \frac{1}{\lambda} \, e^{-x/\lambda} ....
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