Let X and Y be independent random variables such that $X∼Exp(1)$ and $Y∼Exp(2)$. Find the probability that $3X+4Y≤5$.

I thought the equation for solving this was the integral from $\int_0^a \int_ 0^{(a-y)} \lambda_1 e^{-\lambda_1 x}\lambda_2 e^{-\lambda_2 y} dx dy$ where $a = 5$, so I first started with the integral from $\int_0^5 \int_0^{(5-y)} 3 \times 1e^{-1x} \times 4 \times 2e^{-2y} dx dy$ and got $23.83828927$ which is much larger than a probability can be. Any help?

  • 2
    $\begingroup$ There are two issues: your integral evaluates the integral of $3\times 4=12$ over the triangle $X+Y\le 5, X\ge 0, Y\ge 0$ (which is not what the problem asks for). You evaluated it incorrectly; it works out to $6e^{-10}(e^5-1)^2 \approx 5.919$. $\endgroup$
    – whuber
    Apr 29, 2014 at 19:46
  • 2
    $\begingroup$ math.stackexchange.com/q/775500 $\endgroup$
    – Did
    Apr 30, 2014 at 10:41

2 Answers 2


The probability density function of an Exponential($\lambda$) variate is, as shown in the question, given by

$$f_\lambda(x) = \lambda\exp(-\lambda x),\ x\ge 0.$$

($f$ is zero for $x\lt 0$.)

The independence assumption says the probability densities of $X$ and $Y$ multiply to give the joint probability density as

$$f_{X,Y}(x,y) = \left(1\times \exp(-1\times x)\right)\left(2\times \exp(-2\times y)\right) = 2\exp(-x-2y).$$

($f_{X,Y}$ is zero if either of $x$ or $y$ is negative.)

Joint densities give probabilities via integration: to find the probability of an event $E$, you integrate the density over the set $E$:

$${\Pr}_{f_{X,Y}}(E) = \iint_E 2 \exp(-x-2y)dx dy.$$

In this case, by remembering the restrictions $x\ge 0$ and $y\ge 0$, $E$ can be written as $\{(x,y)\ |\ 3x + 4y \le 5, x\ge 0, y\ge 0\}.$ Here is a picture of the graph of $f_{X,Y}$ near this region, with it filled in over $E$ to show how the volume under the graph (the integral) corresponds to the desired probability:


(The gray curves are contours of the graph of the joint density. Because the density does not change when $-x-2y$ is held fixed, these contours are all curves of the form $-x-2y=\text{ constant},$ which are portions of parallel lines all perpendicular to the vector $(-1,-2)$.)

Evidently $E$ is a triangle determined by $x\ge 0$, $y\ge 0$, and $3x+4y\le 5$. If we choose to perform the $x$ integration last, we find that for each $x$ the value of $y$ can range from a low of $0$ to a maximum of $(5-3x)/4,$ deduced by solving the simultaneous equations $3x+4y\le 5$ and $y\ge 0$. It is also evident from the figure and the inequalities that $3x$ cannot exceed $5$, whence $x$ must lie between $0$ and $5/3$. Therefore the desired probability should be computed with an integral like

$${\Pr}(3X+4Y\le 5) = \int_0^{5/3}\int_0^{(5-3x)/4} 2\exp(-x-2y) dy dx .$$

Working it out is elementary and not very instructive (but do it anyway to make sure the result is reasonable this time: partial steps in this direction appear in a duplicate thread on the Math site). The key thing to learn and remember is how to use joint densities to represent probabilities; the rest is part of the rote branch of Calculus. It might also be useful to remember that this kind of problem can be solved from first principles with relatively little work and the help of a simple sketch: you multiply (the marginal densities) and integrate the result over the event in question. It's always a good idea to draw a picture of it. The figure is a good summary of the general procedure.


Check the distribution function of the generalized integer gamma distribution.


(Edit) Also, and answering your question better, check theorem 1 from here http://ac.els-cdn.com/S0047259X97917103/1-s2.0-S0047259X97917103-main.pdf?_tid=8558a38e-cfe5-11e3-b244-00000aacb35d&acdnat=1398807224_cff728a90ab971c59330b2f21843a383 This gives a closed expression for the distribution of what you're looking for.


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