Maths - Solving for the CDF of Y=\sqrt{X}

[Mentallic]

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 textbook 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.

 

[╔(σ_σ)╝]

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.

 

[Mentallic]

Oh, cheers

 

[Ray Vickson]

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 textbook 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

 

[Mentallic]

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.

 

[Ray Vickson]

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

 

[Mentallic]

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]

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.

It all depends on where you start. I am so used to the exponential CDF, F(x) = 1 - exp(-ax), that I write it down almost without thinking. Of course it was obtained way in the past by integration. Still, doing its integral is a little bit easier than integrating your f_Y(y).

RGV