Ok, so I have this written in my notes and while going over it I have a few questions:(adsbygoogle = window.adsbygoogle || []).push({});

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

[tex]

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

[/tex]

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(X^{3}≤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 X^{3}for Y like so:

f_{Y}(y) = P(Y=y) = P(X^{3}=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!

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# Transformation of Random Variables

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