# Limits of Integration - Joint probability density

Given : $$f(x,y)$$ = $$e^{-x-y}$$ if $$x>0 , y>0$$ and $$0$$ elsewhere

Find $$P(X+Y>3)$$ .

For limit I proceeded this problem in the same method this was done Joint probability density function and limits of integration

Sol:
$$P(X+Y>3)=1-P(X+Y<3)$$, now I have tried the integration of $$P(X+Y<3)$$ , $$\int_0^3 \int_0^{3-x} e^{-x-y} dydx \$$ But the integration limits are wrong when I checked , the actual limits are $$\int_2^3\ \int_2^3\ e^{-x-y} dydx$$ ,here they have found directly $$P(X+Y>3)$$

Similarly , for $$f(x,y) = 1/y$$ for $$0 ,find $$P(X+Y>1/2)$$ .Integration part is totally different.

Help Please ! How was the 2 lower limit determined ?

And can any one refer a book where I can learn how to solve probability problems like these. Thanks in advance!

Solution given

The problems are more related to evaluation of multiple integrals than probabilities.

So I suggest taking a look at worked out examples of double/triple integrals in any standard multivariable calculus textbook.

Direct evaluation gives

\begin{align} P(X+Y>3)&=\iint_{x+y>3}e^{-(x+y)}\mathbf1_{x,y>0}\,dx\,dy \\&=\int_0^\infty \left(\int_{\max(0,3-y)}^\infty e^{-x}\,dx\right)\,e^{-y}\,dy \\&=\int_0^\infty e^{-\max(0,3-y)}e^{-y}\,dy \\&=e^{-3}\int_0^3\,dy+\int_3^\infty e^{-y}\,dy \\&=\frac{4}{e^3} \end{align}

• The integration part wasn't the one hard for me, but find the limits were. I am now confused whether the answers given here are right or the ones given in the book [John E. Freund's Mathematical Statistics with Applications 8th Edition (English, Paperback, Irwin Miller, Marylees Miller)] Commented Nov 9, 2018 at 17:28
• Well, finding the limits is part of the integration ;-) Could you mention the page number or perhaps upload the solution given in the book? As your question stands at this moment, I am pretty sure this answer is correct. Commented Nov 9, 2018 at 17:49
• Can you suggest some standard books? and I have uploaded the Screenshots of the problem and solution. Commented Nov 10, 2018 at 8:59
• @DravidHemanth Looks like the solution is to a different problem, not the one you have posted. In any case, it is wrong. They have calculated something like $P(2<X,Y<3)$. Commented Nov 10, 2018 at 9:35

For $$P\{X+Y <3\}$$ the correct expression is $$\int_0^{3}\int_0^{3-x} e^{-x-y}\, dy\, dx$$. note the upper limit $$3-x$$.

• Yes, sir, I have corrected it, but how are 0 to 3-x limits of x? Aren't they limits of y? Commented Nov 9, 2018 at 12:14
• @DravidHemanth Sorry, that was a typo. Anyway there is no way the lower limit can be $2$. Commented Nov 9, 2018 at 12:15

Alternative methods:

1. The given joint pdf is $$f_{XY}(x,y)=f_X(x)f_Y(y)$$ where $$X,Y$$ are iid exponential with mean 1. Thus

$$Z=(X+Y)\sim\text{Gamma}(2;1)$$

and

$$2Z\sim \chi_{(4)}^2$$

concluding:

$$P(X+Y>3)=P(2Z>6)\approx 0.199$$

using chi-square tables

1. being $$Z\sim\text{Gamma}(2;1)$$ where 1 is the rate parameter, we have

$$f_Z(z)=z e^{-z}$$

thus

$$\mathbb{P}[Z>3]=\int_3^{\infty}z e^{-z}dz=4e^{-3}\approx 0.199$$

Note that in method 2) $$f(z)$$ is the convolution of the two marginals, $$X,Y$$