# Probability of the sum of exponential random variables

Let $X$ and $Y$ be independent random variables such that $X∼Exp(1)$ and $Y∼Exp(2)$. Find the probability that $3X+4Y≤5$. Give at least 10 correct digits after the decimal point.

So the formula for $X$ is $e^{-x}$ and the formula for $y$ is $2e^{-2x}$ I believe since this is asking for an exponential with parameters 1 and 2. I'm guessing we need to set up and integral but I'm a little confused on this.Thanks

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Do you know what happens to their distributions when you sum two independent random variables? – Seyhmus Güngören Apr 30 '14 at 10:39
Duplicate of stats.stackexchange.com/questions/95687/… – Juho Kokkala Apr 30 '14 at 10:40

For every $x\gt0$, $[3x+4Y\lt5]=[Y\lt(5-3x)/4]$ hence $P(3x+4Y\lt5)=1-\mathrm e^{-2(5-3x)/4}$ if $x\lt5/3$ and $=0$ otherwise. Thus, $$P(3X+4Y\lt5)=\int_0^{5/3}(1-\mathrm e^{-2(5-3x)/4})\,\mathrm e^{-x}\,\mathrm dx.$$ Surely you can end this.

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may I ask how you came up with this integral to calculate – Jilly Apr 30 '14 at 10:47
Using the value of $P(3x+4Y\lt5)$ recalled just before + the distribution of $X$ + the independence of $X$ and $Y$. – Did Apr 30 '14 at 10:55
Did you downvote by any chance? – Did Apr 30 '14 at 11:39

If rv is exponentially distributed then it can be convenient to work with $1-F(x)$ instead of $F(x)$.

$$P\left[3X+4Y>5\right]=\int_{0}^{\infty}P\left[Y>\frac{5-3x}{4}\right]e^{-x}dx$$ $$=\int_{0}^{\frac{5}{3}}e^{-\frac{5-3x}{2}-x}dx+\int_{\frac{5}{3}}^{\infty}e^{-x}dx=\int_{0}^{\frac{5}{3}}e^{-\frac{5-5x}{4}}dx+P\left[X>\frac{5}{3}\right]$$

and off course:$$P\left[3X+4Y\leq5\right]=1-P\left[3X+4Y>5\right]$$

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