# Continuous Functions of One Random Variable

1. Nov 8, 2012

### SpiffWilkie

My problem is as follows (sorry, but the tags were giving me issues. I tried to make it as readable as possible):
Let X have the pdf f(x)= θ * e-θx, 0 < x < ∞

Find pdf of Y = ex

I have
G(y) = P(X < ln y) = ∫ θ * e-θx dx from 0 to ln y = 1 - y

So then I need g(y) = G'(y) of Y? which would be
θ * e-θln y?

I know there's something obvious in there that I'm missing. I'm having a hard time connecting the last equation to the process that I usually follow.

If anyone can give a nudge, I'd be grateful.

2. Nov 8, 2012

### Mute

Have you taken the derivative of the expression 1-y with respect to y, rather than writing y in exponential form? That will give you your probability density for y. You have a slight mistake in your first attempt θe-θln y - this can be rewritten as θ y; can you spot what the problem is and how it can be fixed?

3. Nov 8, 2012

### SpiffWilkie

As in (θy)/y ?
As for that, am I missing 1/y, so I get (θ y)/y ?

Sorry if I'm not grabbing onto what your trying to say. The more I think about it the muddier I get.

4. Nov 8, 2012

### Ray Vickson

You did not make an error; the result you had --- P(Y < y) = 1 - y^(-θ) --- is correct exactly as written. However, you need to specify the interval {y ≥ 1}, because y = exp(x) ≥ 1 for x ≥ 0. In other words, P{Y < y} = 0 for y < 1.

RGV

5. Nov 9, 2012

### SpiffWilkie

Big, fat facepalm going on over here....

That's why I was so unsure of what I was getting. I was trying to integrate from 0 to infinity to check the PDF, which wasn't giving me 1.
Stupid interval...

Anyhow, thank you very much for the explanation. It helped out a lot. Off to practice some more problems...