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.
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$\begingroup$ If they are 100% correlated that means they are essentially the same so the pdf’s would scale by N $\endgroup$ – Dan Boschen Feb 19 at 13:03
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$\begingroup$ How is the correlation defined? $\endgroup$ – AlexTP Feb 19 at 20:09
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$\begingroup$ Unless you specify the joint PDF of the exponential random variables, your question is not answerable at all. $\endgroup$ – Dilip Sarwate Feb 20 at 1:37
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$\begingroup$ @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. $\endgroup$ – AlexTP Feb 20 at 9:40
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$\begingroup$ @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. $\endgroup$ – Dilip Sarwate Feb 20 at 15:23
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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.
answered Feb 23 at 18:47
