# Exponential Distribution

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## Homework Statement

If we have the exponential distribution $$f_X(x)=\frac{1}{2}e^{-x/2}$$ then show that the cumulative distribution function of $$Y=\sqrt{X}$$ is given by $$F_Y(y)=1-e^{-y^2/2}$$

## Homework Equations

$$F_Y(y)=f_X(x)\cdot\left| \frac{dx}{dy}\right|$$

$$F_Y(y)=f_X(h^{-1}(y))\cdot\left| \frac{d(h^{-1}(y))}{dy}\right|$$

## The Attempt at a Solution

$$Y=\sqrt{X}=h(x)$$

$$\therefore h^{-1}(y)=y^2$$

$$\frac{d(h^{-1}(y))}{dy}=2y$$

$$f_X(h^{-1}(y))=\frac{1}{2}e^{-y^2/2}$$

$\therefore$ after plugging these values into the formula in the relevant equations,

$$F_Y(y)=y\cdot e^{-y^2/2}$$

Which is not what I was meant to show. I only had one example in my text book to go off of and I (from what I can tell) think I applied it correctly to my question, but clearly I haven't. Can someone please guide me in the right direction, and also if you can see anything in my steps that need to be scrutinized, don't be afraid to speak out.

You used an incorrect equation.

$f_Y(y)=f_X(x)\cdot\left| \frac{dx}{dy}\right|$.

What you found was the PDF not the CDF. You have to integrate.

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Oh, cheers

Ray Vickson
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Dearly Missed

## Homework Statement

If we have the exponential distribution $$f_X(x)=\frac{1}{2}e^{-x/2}$$ then show that the cumulative distribution function of $$Y=\sqrt{X}$$ is given by $$F_Y(y)=1-e^{-y^2/2}$$

## Homework Equations

$$F_Y(y)=f_X(x)\cdot\left| \frac{dx}{dy}\right|$$

$$F_Y(y)=f_X(h^{-1}(y))\cdot\left| \frac{d(h^{-1}(y))}{dy}\right|$$

## The Attempt at a Solution

$$Y=\sqrt{X}=h(x)$$

$$\therefore h^{-1}(y)=y^2$$

$$\frac{d(h^{-1}(y))}{dy}=2y$$

$$f_X(h^{-1}(y))=\frac{1}{2}e^{-y^2/2}$$

$\therefore$ after plugging these values into the formula in the relevant equations,

$$F_Y(y)=y\cdot e^{-y^2/2}$$

Which is not what I was meant to show. I only had one example in my text book to go off of and I (from what I can tell) think I applied it correctly to my question, but clearly I haven't. Can someone please guide me in the right direction, and also if you can see anything in my steps that need to be scrutinized, don't be afraid to speak out.

Rather than plugging in formulas, I think it is better to proceed from first principles: $$P\{Y \leq y\} = P\{ \sqrt{X} \leq y \} = P\{ X \leq y^2 \} = \left. (1 - e^{-x/2})\right|_{x=y^2}.$$

RGV

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P\{ X \leq y^2 \} = \left. (1 - e^{-x/2})\right|_{x=y^2}. [/tex]

Doesn't the jump between these two steps defeat the purpose? I believe I'm meant to use the integral to show that.

Now, if I were to calculate P(Y<x) for example, wouldn't I need to integrate again? Which means that I'd need to find a numerical solution.

edit: Never mind, I wouldn't integrate again.

Last edited:
Ray Vickson
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Dearly Missed
Doesn't the jump between these two steps defeat the purpose? I believe I'm meant to use the integral to show that.

Now, if I were to calculate P(Y<x) for example, wouldn't I need to integrate again? Which means that I'd need to find a numerical solution.

edit: Never mind, I wouldn't integrate again.

There is no such a thing as "meant to" (unless stated explicitly). Your are "meant to" do anything that is correct and that you feel comfortable doing; aside from that, there are no rules. If you prefer to use the formulas you wrote before, go ahead and do that, but there are other ways to proceed, too, and I showed you one of them (which happens to be the one I personally prefer).

RGV

Homework Helper
There is no such a thing as "meant to" (unless stated explicitly). Your are "meant to" do anything that is correct and that you feel comfortable doing; aside from that, there are no rules. If you prefer to use the formulas you wrote before, go ahead and do that, but there are other ways to proceed, too, and I showed you one of them (which happens to be the one I personally prefer).

RGV

Sorry, I guess I wasn't clear. What I was trying to say was, didn't you arrive at that answer by integrating? It seems that in between those two steps there was some integration involved. And yes, I've also done essentially what you have there.

Ray Vickson