# Method of Distribution Functions problem

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• Apr 21st 2013, 03:29 PM
Migno
Method of Distribution Functions problem
I need help with this problem:

"The joint distribution of amount of pollutant emmited from a smokestack without cleaning device(Y_1) and a similar smokestack with a cleaning device (Y_2) is

f(y_1,y_2)= 1, 0<=(less or equal to)y_1<=2, 0<=y_2<=1, 2y_2<=y_1
0, elsewhere

U = Y_1 - Y_2

Find the probability density function for U"

So I need to find F_U(u)=P(U<=u)=P(Y_1-Y_2<=u)=P(Y_1<=u+Y_2)

I drew the attachment, but I don't know how to proceed form there. I don't understand how to choose the region U<=u or where u splits.
I know for u<0, F_U(u)=0 since Y_1-Y_2<0 has no solution and for u>2, F_U(u)=1
• Apr 22nd 2013, 05:32 AM
chiro
Re: Method of Distribution Functions problem
Hey Migno.

Hint: Using your information about the joint distribution of Y1 and Y2, where is Y1 - Y2 defined? (This will give you the region for the random variable and will help you get its distribution).
• Apr 22nd 2013, 11:02 AM
Migno
Re: Method of Distribution Functions problem
Well, 0<Y1-Y2<2 so 0<u<2
I guess that's the region for U and if I let Y2=0 I get the intercept u=1, so I guess that's where it splits.
So for 0<u<1 the region is above the line Y1=u+Y2 and for 1<u<2 the region is below that line. Is that right?
• Apr 22nd 2013, 06:20 PM
chiro
Re: Method of Distribution Functions problem
Now you can integrate over that region with respect to a varying value of u (with your limits) to get a CDF and differentiate to get a PDF in terms of u.

Can you write an integral in terms of u given the limits of joint distribution that satisfy Y1 - Y2 being in the set?

(Hint: Think about each value of u as a slice for each valid value of Y1 - Y2 in the integration region and the probability for a particular u sums up all the probabilities in that slice. To sum something up, you integrate the probability for that slice and that becomes your PDF for P(U = u)).
• Apr 23rd 2013, 12:15 AM
Migno
Re: Method of Distribution Functions problem
Ok, so the drawing was bad and it was throwing me off so I attached a better one. So the area Y_1-Y_2<u is always above the line Y_1-Y_2=u and with Y_2=0 we have that u=Y_1. After the line sort of passes the point (2,1) though, the integration limits will not be the same and at that point u=1, so that's why it has to split there.

a) So for 0<u<1 we have the limits 0<y_2<u and 2y_2<y_1<u+y_2 for 1dy_1dy_2. Integrating part is easy so F_U(u)=(u^2)/2

b) and for 1<u<2 the upper area gets complicated to integrate so you take the complement (A in the attachment) the limits are y_2+u<y_1<2 and 0<y_2<2-u again for 1dy_1dy_2. So F_U(u)=1-((2-u)^2)/2, the one minus is from the fact that it is the compliment.

You differentiate those and get the pdf for u.

Thanks chiro for helping me out. I thought I would share this since it took me forever to figure it out. :)