Sum of exponential random variable with different means

Question

Suppose that $X$ and $Y$ are independent exponential random variables with pdf's $f(x)=\lambda e^{-\lambda x}$ and $f(y) = \mu e^{- \mu y}$. What is $\;P \{ X+Y <t \}$ ie what is the cdf of the sum? I know that the distribution is gamma when the parameter is the same, but I'm not sure of a closed form when the parameters are different.

Answer 1

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $\large\ t > 0$: \begin{align} &\color{0000ff}{\large{\rm P}\braces{x + y < t}} \\[3mm]&=\int_{0}^{\infty}\lambda\expo{-\lambda x}\int_{0}^{\infty} \mu\expo{-\mu y}\Theta\pars{t - x - y}\,\dd x\,\dd y = \lambda\mu\int_{0}^{\infty}\expo{-\lambda x}\int_{0}^{\infty} \expo{-\mu y}\Theta\pars{\bracks{t - x} - y}\,\dd y\,\dd x \\[3mm]&= \lambda\mu\int_{0}^{\infty}\expo{-\lambda x}\Theta\pars{t - x}\int^{t - x}_{0} \expo{-\mu y}\,\dd y\,\dd x = \lambda\mu\int_{0}^{\infty}\expo{-\lambda x}\Theta\pars{t - x}\, {\expo{-\mu\pars{t - x}} - 1 \over -\mu}\,\dd x \\[3mm]&= \lambda\int_{0}^{t} \bracks{\expo{-\lambda x} - \expo{-\mu t}\expo{\pars{\mu - \lambda}x}}\,\dd x = \lambda\bracks{% {\expo{-\lambda t} - 1 \over -\lambda} - \expo{-\mu t}{\expo{\pars{\mu -\lambda}t} - 1 \over \mu - \lambda}} \\[3mm]&= \lambda\bracks{% {1 - \expo{-\lambda t} \over \lambda} + {\expo{-\mu t} - \expo{-\lambda t} \over \mu - \lambda}} = \lambda\, {\pars{\mu - \lambda}\pars{1 - \expo{-\lambda t}} + \lambda\pars{\expo{-\mu t} - \expo{-\lambda t}} \over \lambda\pars{\mu -\lambda }} \\[3mm]&= {\pars{\mu - \lambda} - \pars{\mu - \lambda}\expo{-\lambda t} + \lambda\pars{\expo{-\mu t} - \expo{-\lambda t}} \over \mu -\lambda} =\color{#0000ff}{\large% 1 - {\mu\expo{-\lambda t} -\lambda\expo{-\mu t} \over \mu - \lambda}} \end{align}

Answer 2

Felix has provided a very nice detailed solution by hand. This type of problem can also be easily solved by automated methods using a computer algebra system ...

By independence, the joint pdf of $(X,Y)$ is say $f(x,y)$:

We seek the cdf of the sum, i.e. $P(X+Y<t)$:

where I am using the Prob function from the mathStatica add-on to Mathematica to calculate the probability automatically. All done.

Notes:

1. The answer is the same as that obtained by Felix.
2. As formal disclosure, I should perhaps add that I am one of the authors.

Answer 3

There is another way to solve this. By using the Hypoexponential distribution. My answer explaining that distribution more is here.

Sum of exponential random variables with different parameters - followup

I quote my note from that answer which is:

I've used this extensively while trying to calculate the probability of the sum of several Exponential random variables with different rates being greater than some constant. If you ever need to do that, you could use the following:

Let $X_i, i=1..n$ be Exponential random variables with rates $\lambda_i, i = 1.. n$ such that $\lambda_i \ne \lambda_j$ for $i \ne > j$.

$Y := \sum_{i=1}^{n}X_i$

Then,

$f(Y>t) = \sum_{i=1}^{n}C_{i,n}e^{-\lambda_it}$

where

$C_{i,n} = \prod_{j\ne i}\frac{\lambda_j}{\lambda_j-\lambda_i}$

Therefore, the answer to your question would be:

$With \ t > 0$

$f(X+Y < t) = 1 - f(X+Y > t)$

Here:

$f(X + Y >t) = \sum_{i=1}^{n}C_{i,n}e^{-\lambda_it}$

where

$C_{i,n} = \prod_{j\ne i}\frac{\lambda_j}{\lambda_j-\lambda_i}$

Remark: $\lambda_1 = \lambda$ and $\lambda_2 = \mu$ per your question

$f(X+Y > t) = \frac{\mu}{\mu-\lambda} e^{-\lambda t} + \frac{\lambda}{\lambda-\mu} e^{-\mu t} $

$\therefore f(X+Y < t) = 1-f(X+Y > t) = 1 -\frac{\mu}{\mu-\lambda} e^{-\lambda t} - \frac{\lambda}{\lambda-\mu} e^{-\mu t}$

Which is the same as the answer provided by Felix. However, I do prefer the use of the hypoexponential distribution for direct application, in general. It is pretty easy to do by hand as well.