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Transformation of Random Variables

  1. Apr 21, 2012 #1
    Ok, so I have this written in my notes and while going over it I have a few questions:

    Suppose cubical boxes are made so that the length, X (in cm) of an edge is distributed as

    for 9≤X≤11
    0 otherwise

    What sort of distribution will the volume, Y, of the boxes have, Y in cm^3.

    So in my notes it says to do this:

    FY(y) = P(Y≤y) = P(X3≤y)=P(X≤y1/3)=FX(y1/3)

    But why is it not possible to go straight from the PDF of X to the PDF of Y, using the same technique of substituting X3 for Y like so:

    fY(y) = P(Y=y) = P(X3=y)=P(X=y1/3 )=fX(y1/3)

    Have put this here, because it isn't a homework question, more a general question that I've come across while revising but by all means move it if you disagree!
    Last edited: Apr 21, 2012
  2. jcsd
  3. Apr 21, 2012 #2
    Pressed submit by accident, haven't finished writing the OP!

    Edit: Finished now!
    Last edited: Apr 21, 2012
  4. Apr 21, 2012 #3

    Stephen Tashi

    User Avatar
    Science Advisor

    This is good question and an important one. (You see that the two methods produce different answers, right?)

    The PDF of a discrete random variable X can be interpreted as the "the probability that ...", but it is technically incorrect to interpret it this was for a continuous distribution. [itex] f_X(x) [/itex] is not [itex] P(X = x) [/itex]. Often you can get away with thinking of continuous PDF's the wrong way and still get the right formulas. It's rather like how people think of [itex] \frac{dy}{dx} [/itex] as the ratio of two finite numbers and this helps them remember formulas in calculus. Thinking the wrong way is often helpful but it has pitfalls.

    A better way to think is that [itex] f_X(x) [/itex] is a function that is one factor in an expression that approximates the probability for [itex] X [/itex] being in an interval. For example, [itex] P(x - dx \le X \le x + dx) \approx f_X(x) 2 dx [/itex], thinking of [itex] dx [/itex] as a finite length.

    If we approach this problem by reasoning with PDFs, we must use intervals and things don't look simple.

    [tex] P( y - dy \le Y \le y + dy) = P( y - dy \le X^3 \le y+ dy) [/tex]
    [tex]= P( (y-dy)^\frac{1}{3} \le X \le (y+dy)^\frac{1}{3}) [/tex]
    [tex] = \int_{(y-dy)^\frac{1}{3}}^{(y+dy)^\frac{1}{3}} f_X(x) dx [/tex]
    [tex] = \bigg|_{(y-dy)^\frac{1}{3}}^{(y+dy)^\frac{1}{3}} (\frac{1}{2} x) [/tex]

    Perhaps we can do more manipulations with the dy's and dx's to get to the right answer. At least this suggests that, plugging-in [itex] x= y^\frac{1}{3} [/itex] into [itex] f_X( x) [/itex] isn't likely to work.

    The question is related to a question from calculus: When we make the substitution x = g(y) in an integration, why can't we just change dx to dy? Why does the substitution involve a g'(y) ?
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