Thread: Finding distribution of new random variable

1. Finding distribution of new random variable

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.

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

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

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

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

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

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

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

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

3. Normal distribution

Hey really thanks a lot.
I think the case when $\displaystyle F_X(F^{-1}_X (y/2))=y/2$ 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
$\displaystyle F_X(Phi^{-1}_X (y/2))$ when in this case, Y= 2φ(X) rather then Y= 2F(x)
Will the answer be still be $\displaystyle y/2$ ??