Finding distribution of new random variable

**noob mathematician** · March 15th 2009, 07:25 PM

If X is continous with distribution function F(x) and pdf f(x). It is given
Y=2X, finding the distribution of Y will be:
Fy(a)= P(Y< a)
= P(2X< a)
= P(X< a/2)
= Fx(a/2)
Differentiation will give me:
fy(a) = 1/2fx(a/2)

What if the relationship is Y= 2 F(X), where F(X) is the cumulative distribution of X. I can't seem to make X the subject of the above equation.

**matheagle** · March 16th 2009, 05:59 PM

I'll aussme $F_X(x)$ has an inverse.
But we all know what happens when we ass/u/me.

$F_Y(y)=P(Y\le y)=P(2F_X(x)\le y)=P(F_X(x)\le y/2)$

$=P(X\le F^{-1}_X (y/2))=F_X(F^{-1}_X (y/2))=y/2$ .

This seems right and $0<y<2$ .

So $Y\sim U(0,2)$ .

Or $f_Y(y)=f_X(x)|dx/dy|$ .

With $y=2F_X(x)$ we have $dy/dx=2f_X(x)$

giving us $f_Y(y)=f_X(x){1\over 2f_X(x)}=1/2$ .

**noob mathematician** · March 17th 2009, 04:02 AM

Hey really thanks a lot.
I think the case when $F_X(F^{-1}_X (y/2))=y/2$ really help a lot!

By the way, assuming now X is a continous normal distribution with cumulative density φ(X), will this still work? For example may i use
$F_X(Phi^{-1}_X (y/2))$ when in this case, Y= 2φ(X) rather then Y= 2F(x)
Will the answer be still be $y/2$ ??

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