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Thread: Joint Density f(X,Y) integration when there is no Y in the eqn

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    Re: Joint Density f(X,Y) integration when there is no Y in the eqn




    Ah ok, thank you! I'm on to part

    (c) Find E(X), E(Y ), E(Y |X) and V ar(X|Y ).

    I get the following for E(X) which I think is correct:
    E(x)=\int_x xf(x) dx = \int_0^1 xa(1-x) dx = a \int_0^1 1x-x^2 dx = a \left. (1x^2/2 - x^3/3) \right|_0^1 = 1/3

    But when I do it for E(y) I run into a problem on the 6th step:
    E(y)=\int_y yf(y) dy = \int_0^1 ya(1-x) dy = a\int_0^1 y(1-x) dy = a\int_0^1 (y-xy) dy
    = a \left.(y^2/2-xy^2/2) \right|_0^1 = a(1/2 - x1/2 - 0 - 0)

    I don't know what to do with the X in a(1/2 - x1/2 - 0 - 0). I need to solve for a single number for E(y) but the x is giving me problems.

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    Re: Joint Density f(X,Y) integration when there is no Y in the eqn

    You still have not get the correct marginal pdf for Y.

    Note

    f_Y(y) = \int_{-\infty}^{+\infty} f_{X, Y}(x, y) dx= \int_0^1 a(1 - x)dx, 0 < y < 1

    is independent of x, as you have already integrate the dummy variable x out.

    Actually, from the given joint density, without doing the integration you
    can immediately tell Y \sim \mathrm{Uniform}(0, 1)
    and is independent from X.

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    Re: Joint Density f(X,Y) integration when there is no Y in the eqn

    Hmm. How do I have the X value already out? Im not really following.

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    Re: Joint Density f(X,Y) integration when there is no Y in the eqn

    Do you know the uniform distribution?

    Or more directly, can you get the result

    f_Y(y) = 1, 0 < y < 1

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    Re: Joint Density f(X,Y) integration when there is no Y in the eqn

    Ok, firstly, can f(x,y) = a(1-x),  0=<X=<1,  0=<Y=<1 be seen to be uniform before any work is completed?

    Which leads to my next question, how can I deduce that

    Actually, from the given joint density, without doing the integration you
    can immediately tell Y \sim \mathrm{Uniform}(0, 1)
    and is independent from X.
    What is the process to come to that result?

    Lastly, would the marginal distribution of Y then simply be
    f(y) = a(1-x)

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    Re: Joint Density f(X,Y) integration when there is no Y in the eqn

    Quote Originally Posted by crusoe View Post
    Ok, firstly, can f(x,y) = a(1-x),  0=<X=<1,  0=<Y=<1 be seen to be uniform before any work is completed?
    The joint density isn't uniform. The marginal of Y is uniform. The joint depends on the value of x so it can't be uniform.

    Lastly, would the marginal distribution of Y then simply be
    f(y) = a(1-x)
    This can't be a marginal distribution for Y. Why not? Well it's a function of x... and the marginal distribution needs to be a function of y only.

    Edit: Also, you already know what 'a' is. Stop referring to 'a' and instead replace any reference of 'a' with 2.

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    Re: Joint Density f(X,Y) integration when there is no Y in the eqn

    So what would the marginal distribution for Y be then? I think BGM confused me more.

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    Re: Joint Density f(X,Y) integration when there is no Y in the eqn

    lol sorry to confuse you.

    Actually I think that in the process of solving the normalizing constant a,
    you should already have a rough idea about the marginal distributions of
    X and Y respectively.

    Anyway, let see if you agree with the following steps. You are almost got
    everything correct (except some places do not substitute a = 2 back, I do not know why).
    Just reply if you cannot accept/agree with any one of them.

    1. The solution a = 2

    2. Relationship between the joint distribution and marginal distribution:
    f_Y(y) = \int_{-\infty}^{+\infty} f_{X, Y}(x, y) dx

    3. The given joint distribution:
    f_{X, Y}(x, y) = 2(1 - x), 0 < x < 1, 0 < y < 1

    4. Combining 2 and 3 gives
    f_Y(y) = \int_0^1 2(1 - x)dx

    5. Integration of polynomials:
    \int x^n dx = \frac {x^{n+1}} {n + 1} + C, n = 0, 1, 2, ...

    6. Integration is linear:
    \int af(x) + g(x)dx = a\int f(x)dx + \int g(x) dx

    7. Combining 5 and 6 gives
    f_Y(y) = \int_0^1 2(1 - x)dx = 2\int_0^1 dx - 2\int_0^1 xdx= \left. 2\left(x - \frac {x^2} {2}\right)\right|^{x=1}_{x=0}

    = 2\left(1 - \frac {1^2} {2}\right) - 2\left(0 - \frac {0^2} {2}\right)= 2 \times \frac {1} {2} - 2 \times 0 = 1, 0 < y < 1

    8. So the conclusion is
    f_Y(y) = 1, 0 < y < 1 and Y \sim \mathrm{Uniform}(0, 1)

    If you are familiar with the definition of independence,
    then you should know X, Y are independent if and only if
    f_{X, Y}(x, y) = f_X(x)f_Y(y) ~ \forall x, y
    provided that their joint and marginal density exists.

    So by observations, if the joint density can be written as
    f_{X, Y}(x, y) = g(x)h(y) for some functions
    g(x) independent of y and h(y) independent of x,
    (of course the support is somehow "rectangular", independent of x, y,
    e.g. [0, 1] \times [0, 1] in this case)
    then we know that g(x) = cf_X(x), h(y) = \frac {1} {c} f_Y(y) ~\exists c \neq 0
    and X, Y are independent.

    Thats the reason I say that by observation, without really doing integration,
    you should be confident to say they are independent. Also, since the joint
    distribution is independent from the dummy variable of Y,
    the marginal pdf of Y is a constant function (on [0, 1]),
    so the marginal distribution of Y must be a uniform distribution.

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    Re: Joint Density f(X,Y) integration when there is no Y in the eqn

    That all makes sense, thank you. So back to my question regarding E(Y)

    (c) Find E(X), E(Y ), E(Y |X) and V ar(X|Y ).

    I get the following for E(X) which I think is correct:

    E(x)=\int_x xf(x) dx = \int_0^1 xa(1-x) dx = a \int_0^1 1x-x^2 dx = a \left. (1x^2/2 - x^3/3) \right|_0^1 = 1/3
    But when I do it for E(y) I run into a problem on the 6th step:
    E(y)=\int_y yf(y) dy = \int_0^1 ya(1-x) dy = a\int_0^1 y(1-x) dy = a\int_0^1 (y-xy) dy
    = a \left.(y^2/2-xy^2/2) \right|_0^1 = a(1/2 - x1/2 - 0 - 0)

    I don't know what to do with the X in a(1/2 - x1/2 - 0 - 0). I need to solve for a single number for E(y) but the x is giving me problems.

  10. #25
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    Re: Joint Density f(X,Y) integration when there is no Y in the eqn


    Ok. So back to the problem.

    Do you agreed that

    1. f_Y(y) = 1, 0 < y < 1

    2. E[Y] = \int_{-\infty}^{+\infty} yf_Y(y)dy

    3. E[Y] = \int_0^1 ydy

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