# Sum of correlated exponential random variables

How do we find PDF of sum of correlated exponential random variables. I know for independent random variables. But how to find it for correlated exponential random variables.

• If they are 100% correlated that means they are essentially the same so the pdf’s would scale by N – Dan Boschen Feb 19 at 13:03
• How is the correlation defined? – AlexTP Feb 19 at 20:09
• Unless you specify the joint PDF of the exponential random variables, your question is not answerable at all. – Dilip Sarwate Feb 20 at 1:37
• @DilipSarwate as an exponential random variable i fully characterized by its mean, I believe if the correlation is defined, we can derive the joint PDF. For example, if we know the transform function $Y=g(X)$ then we know the joint pdf. The fact that both $X$ and $Y$ are exponential makes the calculation easier. Of course, the knowledge about $g(.)$ is crucial. – AlexTP Feb 20 at 9:40
• @AlexTP Two exponential random variables given means and with specified correlation coefficient can nonetheless have infinitely many different joint pdfs. Furthermore, if $Y=g(X)$ with $X$ an exponential random variable, then $Y$ is not an exponential random variable (as it must be as per the requirements in the problem statement) except when $g$ is a linear function ($g(x) = ax$ with $a > 0$) in which case the correlation coefficient is $1$. Please rethink your comment, and possibly give some thought to deleting it entirely. If you choose to delete, I will delete this response too. – Dilip Sarwate Feb 20 at 15:23

The pdf $$f_Z(z)$$ of the sum $$Z=X+Y$$ of any two jointly continuous random variables $$X$$ and $$Y$$ with joint pdf $$f_{X,Y}(x,y)$$ is as follows: $$\text{For all } z, -\infty < z < \infty, ~~ f_Z(z) = \int_{-\infty}^\infty f_{X,Y}(x,z-x) \, \mathrm dx.\tag{1}$$
For the special case when $$X$$ and $$Y$$ are nonnegative random variables (including as a special case, exponential random variables) and so take on nonnegative values only, $$f_{X,Y}(x,y)$$ has value $$0$$ if at least one of $$x$$ and $$y$$ is smaller than $$0$$. Hence, in this case, the integrand $$f_{X,Y}(x,z-x)$$ in $$(1)$$ has value $$0$$ if $$x < 0$$ or if $$z < x$$. Consequently, if $$z$$ is a negative number, then the integrand in $$(1)$$ is always $$0$$ regardless of the value of $$x$$ and therefore so is the integral. All of which is just a long-winded way of saying that $$f_Z(z)$$ has value $$0$$ when $$z<0$$, that is, $$Z$$ takes on nonegative values only, which any idiot could have deduced from the fact that $$Z=X+Y$$ and both $$X$$ and $$Y$$ are nonnegative. But the approach is useful even for $$z>0$$ since now we have that the integrand in $$(1)$$ is zero when $$x<0$$ or when $$x >z$$ and so for nonnegative $$X$$ and $$Y$$, we can simplify $$(1)$$ to $$f_Z(z) = \begin{cases}\displaystyle\int_0^z f_{X,Y}(x,z-x) \, \mathrm dx, & z \geq 0,\\\quad\\ 0, & z < 0\end{cases} \tag{2}$$
No further simplification of $$(2)$$ is possible in general.
For the special case when $$X$$ and $$Y$$ are independent random variables, $$f_{X,Y}(x,y)$$ factors into $$f_X(x)f_Y(y)$$ and so $$(1)$$ becomes the familiar convolution integral and $$(2)$$ the somewhat-less-familiar convolution integral for causal signals. But no such simplification is possible for nonindependent random variables $$X$$ and $$Y$$; we need the joint pdf to calculate $$f_Z(z)$$ and just knowing that $$X$$ and $$Y$$ are correlated random variables (whether exponential or Gaussian or whatever) is not enough.