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SoA released problem #125

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  • SoA released problem #125

    Given f(x)=2x for 0<x<1
    Also it says f(y)=?? for 0<y<x

    Since they say F(x) is the marginal density, a instance of cdf (I'm very vague on this terminology); we can get the pdf by differentiating the marginal density f(x)=2x. Such that F(x)= (2x)' = 2 = F(x,y)

    Ok, with magic I found some formulas:

    F(x,y) = f(x|y)f(y) and
    F(x,y) = f(y|x)f(x)

    f(y) = integral of F(x,y) with respect to x; from the range of x from [?,?]

    Determine the density of X, given Y=y means pdf of X given Y=y OR f(x|Y=y) = f(x|y) = F(x,y) / f(y) = 2/(2-2y) = 1/(1-y)

    ================================================
    ok, so my problem is how do you know the range of x is [y,1] and not [0,1]. Are you suppose to always ASSUME 0<x<1 and 0<y<x MUST COMBINE to be 0<y<x<1 AND you MUST change the ranges to 0<y<x and y<x<1 therefor changing the ranges of x to [y,1]. That's the only way it makes sense.

    Ok, i think i'm just venting that I got fooled by the given range of x (marginal density range) which was clearly meant to trick you into using the wrong range for the pdf later. It's funny because when I started typing this up, I was clueless how the answer came about; then after typing it up and thinking about it more clearly... the answer just seems plain as paint.

    I'm fairly new at this stuff, so feel free to correct any vocabulary I may have wrong.
    P FM MFE C

  • #2
    You aren't really changing the range of x. x has the same range. However the range of y has x invloved in it. What you are doing with that type of problem is determining the area that "plays".

    Draw the picture. Since you are new I will do basics. Whenever you have an inequality, draw the equality. In thise case y=x, shade the side that you need determined by < or >. Be sure to cut off both axes by their max. Here I think it said x<1, so a vertical line at x=1. Y could potentially go to infinity if we were not given a finite value for x. Since y is less than x and x and less than one, y < 1. Like in the Mr hanky song, "Therefore bicariously he loves you". This is called determining the 2 d region. Now, that region we have marked off (by the info given on the right of the problem) has a "density" (given by the problem info on the left). What the density does is makes it where we cant come up with a probability greater than one.

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