I am reviewing some notes from an old statistics course in preparation for a big upcoming exam I have. I have an old book by John E. Freund, Mathematical Statistics (5th Edition) that has a number of problems in it that I'm using to prepare. One of them, 3.67 (pg. 123) asks "Find the joint probability density function of the two random variables $X$ and $Y$ whose joint distribution function is given by:

$F(x,y)=1-e^{-x}-e^{-y}+e^{-x-y}$ for $x>0, y>0$ and $0$ otherwise.

and use the joint pdf to find $P(X+Y>3).$

I thought this was a relatively straight forward problem and I carried out all the calculation, but my final answer $4e^{-4}$ does not coincide with the answer presented in the back of the book (on pg. 641), $(e^{-2} - e^{-3})^2$. I am hoping someone can help me understand what I did wrong (or possibly let me know if the book's answer is wrong). Here is what I did to solve the problem.

  1. Calculate the joint PDF: ${{\partial}^2\over{{\partial x}{\partial y}}}F_{X,Y}(x,y) = e^{-(x+y)}$ for $x>0, y>0$ and $0$ otherwise.

  2. Sketech out the area of integration which is essentially the area above and to the right of the line bounded by the line $y = 3-x$ (in the first quadrant of the Cartesian plane). In other words the area bounded below by the line $y=3-x$ from $0<x<3$ and $y=0$ from $x>3$ and bounded on the left by the $x$ axis.

  3. Since the total area under the PDF must be equal to one, I decided to simply calculate the area under $y= 3-x$ (and bounded by the $x$ and $y$ axis) and subtract that result from 1.

  4. Perform the integration: $1-\int_0^3\int_0^{3-x}e^{-(x+y)}dydx = 1 - (-4e^{-3}+1)=4e^{-3}\approx0.1991483.$

My answer, $0.1991483$, is not even close to the author's answers of $(e^{-2} - e^{-3})^2=0.007318497$

Can someone help me with this and double-check my work? Where did I possibly go wrong? Thanks.

  • 1
    $\begingroup$ Your solution $4e^{-3}$ is correct. $\endgroup$
    – nth
    Apr 9, 2017 at 20:26
  • 1
    $\begingroup$ I would guess that the problem might originally have asked to find the chance that both $X$ and $Y$ lie between $2$ and $3$; that the problem statement was changed; but the answer was not changed. $\endgroup$
    – whuber
    Apr 9, 2017 at 20:30
  • $\begingroup$ Thanks, guys! I was driving myself crazy going over this problem again and again. I suspected the author's answer was incorrect for this problem. I suspect that the answer may have been provided to another problem, problem 3.64 in the book. I'm going to verify now, and I'll post back. much appreciated! $\endgroup$ Apr 9, 2017 at 20:54
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    $\begingroup$ You can easily see that @whuber's assertion is correct, because $F(x,y)=(1-e^{-x})(1-e^{-y})$, i.e. it is the joint distribution of two independent exponentials. $\endgroup$
    – nth
    Apr 9, 2017 at 21:47
  • $\begingroup$ As an aside, I just took a look at the second edition of the book. The question above has changed slightly, but essentially asks the student to calculate $P(X+Y <3)$ on page 143 question 1.6. The author indeed has corrected the answer to be $1-4e^{-3}$. $\endgroup$ Apr 11, 2017 at 18:05

1 Answer 1


The following simulation in R indicates that your solution is correct:

> n <- 1e+6
> x <- rexp(n)
> y <- rexp(n)
> binom.test(sum(x+y>3), n, 4*exp(-3))

    Exact binomial test

data:  sum(x + y > 3) and n
number of successes = 199110, number of trials = 1e+06, p-value = 0.9172
alternative hypothesis: true probability of success is not equal to 0.1991483
95 percent confidence interval:
 0.1983237 0.1998900
sample estimates:
probability of success 
  • 1
    $\begingroup$ Thank you, Jarle. I hadn't though of verifying the calculations this way. $\endgroup$ Apr 9, 2017 at 21:09
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    $\begingroup$ @analyst1 While I probably wouldn't have formally tested it, such simulations are a very handy tool for checking whether your calculated answers are plausible. I use simulation as a check on my algebra very frequently (and often use algebra, where feasible, such as on simple cases or asymptotic or approximate results, as a way of checking that my simulations are not implausible, though often plausiblility checks are more immediately available). It would be unusual to make errors in both that gave the same outcome unless you had misunderstood the problem. It's a good habit to be in $\endgroup$
    – Glen_b
    Apr 10, 2017 at 4:05
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    $\begingroup$ Thank you @Glen_b. This is very useful. I really love learning from experts like you guys about these methods. It's exciting to see how real statistics is done. I'm getting more experience every day, and after I pass this very big exam coming up, I'll be able to get a bit more hands on experience as I will complete my course of study! Thank you again for the helpful tips. $\endgroup$ Apr 10, 2017 at 5:23

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