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Math Help - Probability Density

  1. #1
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    Probability Density

    If the probability density of X is given by
    f(x) = {1+x for -1< x <= 0
    1-x for 0 < x < 1
    0 elsewhere
    and U = X and V = X^2, show that (a) cov(U,V) = 0. (b) U and V are dependent.

    I don't really get this stuff at all. Please help!!
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  2. #2
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    Quote Originally Posted by TheHolly View Post
    If the probability density of X is given by
    f(x) = {1+x for -1< x <= 0
    1-x for 0 < x < 1
    0 elsewhere
    and U = X and V = X^2, show that (a) cov(U,V) = 0. (b) U and V are dependent.

    I don't really get this stuff at all. Please help!!
    If non-one replies in the meantime, I might have time later tonight (my time) to have a closer look.
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  3. #3
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    I would think that you would use the transformation theorem to get the pdf of  V . Its a monotonic , strictly increasing transformation.

    Then show what  p_{X,Y}(x,y) \neq p_{X}(x) \cdot p_{Y}(y) for dependence. And use definition of covariance:  \text{Cov}(X,Y) = E[XY] - \mu_{X} \mu_{Y} .
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  4. #4
    Oli
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    cov(U,V)=E((U-u)(V-v)) where u,v are E(U),E(V) respectfully.

    u=E(U)=E(X)=0, for obvious reasons.

    So we have:

    cov(U,V)=E(U(V-v))
    =E(UV-Uv)
    =E(UV)-E(Uv)
    =E(X^3)-vE(U)
    =E(X^3)-vu
    =E(X^3)-v.0
    =E(X^3)
    =0

    this last step can be done: x^3 is an odd function, f(x) is an even function, so when we multiply them we get an odd function. Hence the integral of x^3.f(x) (ie the expected value of X^3) is 0.

    Dependence follows...
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  5. #5
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    Quote Originally Posted by TheHolly View Post
    If the probability density of X is given by
    f(x) = {1+x for -1< x <= 0
    1-x for 0 < x < 1
    0 elsewhere
    and U = X and V = X^2, show that (a) cov(U,V) = 0. (b) U and V are dependent.

    I don't really get this stuff at all. Please help!!
    (b) Clearly U and V are dependent, by definition, since one variable is determined by the other.

    (a) You need to find \text{cov} (X, X^2) = E(X \cdot X^2) - E(X) \cdot E(X^2) = E(X^3) - E(X) \cdot E(X^2).
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  6. #6
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    Quote Originally Posted by Oli View Post
    cov(U,V)=E((U-u)(V-v)) where u,v are E(U),E(V) respectfully.

    u=E(U)=E(X)=0, for obvious reasons.

    So we have:

    cov(U,V)=E(U(V-v))
    =E(UV-Uv)
    =E(UV)-E(Uv)
    =E(X^3)-vE(U)
    =E(X^3)-vu
    =E(X^3)-v.0
    =E(X^3)
    =0

    this last step can be done: x^3 is an odd function, f(x) is an even function, so when we multiply them we get an odd function. Hence the integral of x^3.f(x) (ie the expected value of X^3) is 0.

    Dependence follows...
    NB: Independence implies covariance = 0 but covariance = 0 does NOT imply independence.
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  7. #7
    Oli
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    Let A be the event U is in (0,0.5)
    Let B be the event V is in (0,0.25).

    P(A)P(B) =/= P(A,B)
    (just do the sums and this is clear: you don't have to do all the sums, just enough to show this is true...)

    This implies dependence.
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  8. #8
    Oli
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    Sorry, when I said dependence follows, I meant I would be adding a proof of dependence afterwards... done it now.

    This is in fact a counter example: the point of the question is to show dependent variables with covariance 0.
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  9. #9
    Oli
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    Some helpful hints with this stuff:

    Expected values are one of your most useful tools with probability theory:
    Know stuff like:
    A random variable can never equal an expected value unless the random variable is constant. So if you end up proving something like:
    E(X)=Y where X,Y are random variables, you are probably wrong.

    Expected value satisfies:
    E(aX+bY)=aE(X)+bE(Y)

    It DOES NOT SATISFY E(XY)=E(X)E(Y) unless X,Y are independent.

    If A is an event, and 1{A} is the indicator function of the event (i.e 1{A} = 1 if event happens, 0 otherwise), then:
    E(1{A})=P(A)
    This is slightly more advanced, but can often be a useful trick.
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