Continuous Functions of One Random Variable

[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 = e^x


I've gone about this the way I normally do for these problems.
I have
G(y) = P(X < ln y) = ∫ θ * e^-θx dx from 0 to ln y = 1 - y^-θ


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.

 

[Mute]

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 = e^xI've gone about this the way I normally do for these problems.
I have
G(y) = P(X < ln y) = ∫ θ * e^-θx dx from 0 to ln y = 1 - y^-θSo 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.

Have you taken the derivative of the expression 1-y^-θ with respect to y, rather than writing y^-θ 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^-θ; can you spot what the problem is and how it can be fixed?

 

[SpiffWilkie]

Have you taken the derivative of the expression 1-y^-θ with respect to y, rather than writing y^-θ in exponential form?

As in (θy^-θ)/y ?

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^-θ; can you spot what the problem is and how it can be fixed?

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

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

 

[Ray Vickson]

As in (θy^-θ)/y ?

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

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

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

 

[SpiffWilkie]

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

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...