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Math Help - The Prbability Density function of the minimum of two random variables

  1. #1
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    The Prbability Density function of the minimum of two random variables

    I am trying to show (or am I even trying to show something that is not valid??) that for arbitrary continuous random variables X and Y, we have

    <br />
F_{min\{X,Y\}}(x) = F_{X}(x) + F_{Y}(x)-\mathbb{P}(X\leq x,Y\leq y),<br />

    where F_{X} is the distriubtion function of the random variable X.


    What I have done is that by using the definition F_{X}:=\mathbb{P}(X\leq x),

    <br />
F_{min\{X,Y\}}(x) &= 1 - \mathbb{P}(min\{X,Y\}\leq x)\\<br />
&= 1 -  \mathbb{P}(X > x , Y > x)\\<br />

    But it seems that this leads me nowhere really. If I assume X and Y are independent, I get the result, but what if they are not?
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  2. #2
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    I dont think you need a fancy proof here.

    P(min(x,y) \leq k) = P(x \leq k ~OR~ y \leq k ~OR~ Both)

    P(min(x,y) \leq k) = P(x< k)  + P(Y \leq k) - P(x \leq k ~ and ~ Y \leq k)

    F_{(min(x,y)}(k) = F_X(k) + F_Y(k) -P(X \leq k,Y \leq k)


    I would think that the second line is considered elementary and you dont have to prove it each time you use it.



    if that doesn't convince you, here is a more elaborate proof for the case of continuous variables.
    Spoiler:

    Define Z=min(x,y)

    Without loss of generality
    , suppose x,y are supported on the range [0,1].


    P(Z \leq z) = \int_{x=0}^{x=z}  \int_{y=x}^{y=1} f_{x,y} ~dx dy ~~ +  \int_{y=0}^{y=z}  \int_{x=y}^{x=1} f_{x,y} ~dy dx
    interpretation: This is the chance that X is the smallest and less than z, plus the chance that Y is the smallest and less than z

    P(Z \leq z) = \left( \int_{x=0}^{x=z}  \int_{y=0}^{y=1} f_{x,y} ~dx dy  - \int_{x=0}^{x=z}  \int_{y=0}^{y=x} f_{x,y} ~dx dy \right)~~ +  \left(\int_{y=0}^{y=z}  \int_{x=0}^{x=1} f_{x,y} ~dy dx  - \int_{y=0}^{y=z}  \int_{x=0}^{x=y} f_{x,y} ~dy dx\right)


    P(Z \leq z) = \left(F_X(z) - \int_{x=0}^{x=z}  \int_{y=0}^{y=x} f_{x,y} ~dx dy \right)~~ +  \left(F_Y(z) - \int_{y=0}^{y=z}  \int_{x=0}^{x=y} f_{x,y} ~dy dx\right)

    P(Z  \leq z) = \left(F_X(z) + F_Y(z) \right) - \left( \int_{x=0}^{x=z}  \int_{y=0}^{y=x} f_{x,y} ~dx dy  + \int_{y=0}^{y=z}  \int_{x=0}^{x=y} f_{x,y} ~dy dx\right)
    Consider the integrals on the right hand side, and the square with points at (0,0) and (z,z). The square can be divided into two triangles by drawing a line from (0,0) to (z,z). The first integral covers the lower triangle, and the second integral covers the upper triangle. Therefore, together they are integrating over the whole square. In the more general case where the support of each variable was not (0,1), the same argument would apply - except that we would be dividing an arbitrary rectangle instead of a square

    The Prbability Density function of the minimum of two random variables-int2.jpg


    P(Z \leq z) = \left(F_X(z) + F_Y(z) \right) - \left( \int_{x=0}^{x=z}  \int_{y=0}^{y=z} f_{x,y} ~dx dy \right)



    P(Z \leq z) = \left(F_X(z) + F_Y(z) \right) - P(X \leq z,Y \leq z)



    Last edited by SpringFan25; June 27th 2010 at 01:23 AM.
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  3. #3
    Moo
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    If they're not independent, it's not easy
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  4. #4
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    i didn't (knowingly) assume independance in either of the proofs i gave, did i miss something?
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  5. #5
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    Yes, the question he asked at the end of his first post
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