How to Calculate the CDF of e^-y Using Double Integrals?

[cse63146]

Homework Statement

$$\int^x_0\int^y_x e^{-v} dv du$$

where u and v are just "dummy variables"

Homework Equations

The Attempt at a Solution

$$\int^x_0\int^y_x e^{-v} dv du = \int^x_0 -e^{-y} + e^{-x} du = (-e^{-y} + e^{-x})x$$

have I made a mistake somewhere?

 

[djeitnstine]

Everything looks fine don't forget your integration constant after ;)

 

[cse63146]

It's for determining a CDF. I don't think I need the constant.

A classmate got x(1 - e^-y). Did they use somesort of trick to get it, or is their answer just wrong?

 

[djeitnstine]

That's very wrong. First of all there has to be at least 2 exponentials, perhaps they just forgot their $$e^{-x}$$

 

[cse63146]

Well, that cleared that up.

Thanks.

 

[cse63146]

Got another question.

$$\int^x_0\int^y_x e^{-v} dv du$$ This case is when 0<x<y

If I want to do 0 < y < x, would the double integral be set up in the following way:

$$\int^x_0\int^x_y e^{-v} dv du$$ ?

 

[djeitnstine]

Hmm before I go into any discussion, are you integrating backwards?? Or are you trying to change the order of integration? If you are changing the order I would say this:

Spoiler

You're changing the order of integration. Let's say your limits are $$y=x^2$$ from 0 < x 6 bounded by y=0 and you are doing the dy dx order or integration then you would have to do
$$\int^{6}_{0} \int^{x^2}_{0} dy dx$$ (from top to bottom, left to right)

Now say you want to change the order, that is, from dx dy, you would have to write y in terms of x and do your integration that way, i.e.

$$\int^{36}_{0} \int^{6}_{\sqrt{x}} dx dy$$ (from left to right, bottom to top)

if not just ignore it and let me know what you mean

 

[cse63146]

I'm trying to find the CDF of e^-y.

The cdf of any f(x,y) is $$\int^x_{-\infty} \int^y_{-\infty} f(u,v) dv du$$

where u and v are just dummy variables. I need to consider 2 cases: 1) 0<x<y and 2) 0 < y < x. In other words I need to do 2 double integrals using the general set up of a CDF.