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Probability- Exponential Distribution/Joint pdfs

  1. Jan 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Let X1, X2 be exponential RVs with parameter λ, μ respectively. (The question does NOT say that they are independent but I think this must be a typo?! If not I have even less idea how to do the question).

    Let Y= min {X1, X2}
    and Z = max {X1, X2}

    Let W = Z-Y
    Calculate the joint pdfs of (Y,Z), (Z,W), (Y,W). Which pairs are independent.


    2. Relevant equations

    I know how to find the the pdf of Y, Z seperately (via the cdf) but this doesn't seem to be directly relevant. Clearly Y,Z are not independent, so I think I need to find their joint pdf by first finding the joint cdf and then integrating.


    3. The attempt at a solution

    So far I have:
    FY,Z(y,z) = P( Y < y, Z < z)
    = P(Y,Z<y) + P(Y<y, y<Z<z)
    = P(X1<y, X2<y) + P(X1 <y, y<X2<z) + P(X2<y, y<X1<z)

    I really would like to know if this line is valid.. from here I used independence of Y,Z to find that
    FY,Z(y,z) = 1 - exp(-y(λ+μ)) - exp(-λz) - exp(-μz) + exp(-μy-λz) + exp(-λy-μz)
    for y<= z.

    so
    fY,Z(y,z) = 0 for y<z
    = λμ(exp(-μy-λz) + exp(-λy-μz)) otherwise

    This looks promising on the basis that the integral first wrt y from 0 to z, and then wrt z from 0 to infinity is 1.

    But if I attempt to find the marginal distribution by integrating from 0 to ∞ wrt z, I get:
    μexp(-μy) + λexp(-λy)
    which is not what I would expect- I think the answer should be
    (μ+λ)exp(-(μ+λ)y)

    Am I even on the right lines at all? Its a much longer question but if this is wrong (as I fear it might be) there's not much point trying to plow through the rest using the wrong formulas..
    Thank you!
     
  2. jcsd
  3. Jan 23, 2012 #2

    Ray Vickson

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    I don't think your expressions for the joint distribution of (Y,Z) are correct. In fact, P{y < Y & Z < z} = P{y < X1 < z & y < X2 < z}.

    RGV
     
  4. Jan 23, 2012 #3
    Ok.. well better to find out its wrong now as I said :P
    Erm.. slightly confused by what you wrote, I think I agree that
    P{y < Y & Z < z} = P{y < X1 < z & y < X2 < z}.

    but seeing as Y,Z are not independent I then don't understand how I can relate this expression to P(Y<y, Z<z). Am I being incredibly slow?
     
  5. Jan 23, 2012 #4

    Ray Vickson

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    If X1 and X2 are independent, you can easily compute P{y < X1 < z & y < X2 < z}.

    RGV
     
  6. Jan 23, 2012 #5
    Yes I know that. My question was I don't understand the relationship between

    P{y < Y & Z < z} and P(Y<y & Z<z)

    Obviously P(y<Y) = 1 - P(Y<y) but the dependence of Y,Z means I can't use this..
     
  7. Jan 23, 2012 #6

    Ray Vickson

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    Your statement "Obviously P(y<Y) = 1 - P(Y<y)" is clearly wrong: we don't have A = 1-A for A = P(Y < y), as you are claiming.

    I don't see your problem. If y = 3 and z = 5, are the numbers 3 and 5 dependent? Of course, Y and Z are dependent, but that is why I was careful to distinguish between y and Y and between z and Z.

    RGV
     
  8. Jan 23, 2012 #7
    Thats not what I wrote-
    P(Y≥y) + P(Y<y) = 1
    last time I checked... my notation made it confusing apologies- the capitals were the other way around.

    However I'm definitely still missing something here.

    P(y < Y & Z < z) = P(Y>y, Z<z) ≠ P(Y<y & Z<z) = FY,Z(y,z)

    So knowing the thing on the left does not help me find the thing on the right.
     
  9. Jan 23, 2012 #8

    Ray Vickson

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    Knowing P(Y>y & Z <z) is enough to determine the joint density f(y,z), but if I tell you how, I would be doing your homework.

    RGV
     
  10. Jan 24, 2012 #9
    Its actually a tiny part of a very long question that I'm taking as part of an optional course. So you're hardly doing my homework. But anyway. Thanks.
     
  11. Jan 24, 2012 #10

    Ray Vickson

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    Sigh. I'll give it one more try. If you _did_ know the bivariate density f(u,v) of (Y,Z) at all relevant values of Y=u and Z=v, you could obtain P(Y>y & Z<x) as a two-dimensional integral of f. Write out this integral. Now it should be obvious how you can get f.

    RGV
     
  12. Jan 24, 2012 #11

    Ray Vickson

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    Sorry: I mis-read your expression for f(y,z) (recovering from a cold/flu and not at my best): your expression _is_ correct, but I thought the derivation more convoluted than need be. Much easier: {Y=y,Z=z} = {X1=y & X2 = z} or {X2=y & x1=z}. (Also, you would get this from the expression H(y,z) = P(Y>y,Z<z), using f(y,z) = -∂^2 H/ ∂y ∂z.)

    Also: your marginal of Y was computed incorrectly: you should integrate over z from z = y to z = ∞.

    RGV
     
  13. Jan 24, 2012 #12
    (Also, you would get this from the expression H(y,z) = P(Y>y,Z<z), using f(y,z) = -∂^2 H/ ∂y ∂z.)

    Yes I had actually worked that out, but I was just about ready to give up when it gave me the same answer I got the first time. Thanks for all your help, I now have some chance of finishing the sheet in time :)
     
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