Transformation of Random Variables

[jimbobian]

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

$$f(x)=\frac{1}{2}$$
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:

F_Y(y) = P(Y≤y) = P(X³≤y)=P(X≤y^1/3)=F_X(y^1/3)

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:

f_Y(y) = P(Y=y) = P(X³=y)=P(X=y^1/3 )=f_X(y^1/3)

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!

 

[jimbobian]

Pressed submit by accident, haven't finished writing the OP!

Edit: Finished now!

 

[Stephen Tashi]

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 !

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. $f_X(x)$ is not $P(X = x)$. 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 $\frac{dy}{dx}$ 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 $f_X(x)$ is a function that is one factor in an expression that approximates the probability for $X$ being in an interval. For example, $P(x - dx \le X \le x + dx) \approx f_X(x) 2 dx$, thinking of $dx$ as a finite length.

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

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

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 $x= y^\frac{1}{3}$ into $f_X( x)$ 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) ?