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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<x<y , 0<y<1 $ ,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!

Question

Solution given

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3 Answers 3

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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}

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  • $\begingroup$ 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)] $\endgroup$ Commented Nov 9, 2018 at 17:28
  • $\begingroup$ 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. $\endgroup$ Commented Nov 9, 2018 at 17:49
  • $\begingroup$ Can you suggest some standard books? and I have uploaded the Screenshots of the problem and solution. $\endgroup$ Commented Nov 10, 2018 at 8:59
  • $\begingroup$ @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 $P(2<X,Y<3)$. $\endgroup$ Commented Nov 10, 2018 at 9:35
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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 $3-x$.

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

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