1

I have 2 exponential distribution with the following probability density function:-

$f(x)= \lambda_1 e^{-\lambda_1 x}$

$g(y)= \lambda_2 e^{-\lambda_2 y}$

And their correlation is given as $\rho$. How can I find/construct a joint probability density function $h(x,y)$ which is consistent to the above?

Many thanks!

H42
  • 165
  • You could (at least for positive correlation) make a mixture of independent variables and a totally dependent pair ($y=ax$). Related :https://math.stackexchange.com/questions/2066774/generating-correlated-arbitrary-random-variables – leonbloy Jun 17 '18 at 16:25
  • Thanks for reply. I just reviewed the post but the respective marginal distributions are referring to Chi-squared and Normal distributions. How can we apply similar techniques to exponential distribution? – H42 Jun 17 '18 at 16:42
  • And what are the domains of f(x) and g(y)? – callculus42 Jun 17 '18 at 18:59
  • @callculus Hi, that's a simple exponential distribution and thus from zero to infinity. Thanks. – H42 Jun 17 '18 at 19:08

1 Answers1

1

Start from $(X_1,Z_1,B)$ independent, $X_1$ and $Z_1$ both standard exponential and $B$ Bernoulli with $P(B=1)=b$, $P(B=0)=1-b$ for some $b$ in $(0,1)$, and define $$Y_1=BX_1+(1-B)Z_1$$ Then $Y_1$ is standard exponential and, by a direct computation, $$\mathrm{Corr}(X_1,Y_1)=E(X_1Y_1)-1=b$$ thus $$X=X_1/\lambda_1\qquad Y=Y_1/\lambda_2$$ fit the bill for every desired nonnegative correlation $b$.

A construction to get some, but not all, negative correlations starts from the observation that, if $X_1$ is standard exponential then $e^{-X_1}$ and $1-e^{-X_1}$ are both uniformly distributed on $(0,1)$ hence $$Y_2=BX_1-(1-B)\log(1-e^{-X_1})$$ is standard exponential. Furthermore, introducing $U$ uniform on $(0,1)$, one gets $$\mathrm{Corr}(X_1,Y_2)=2b-(1-b)E(\log U\log(1-U))-1=2b-(1-b)(2-\pi^2/6)-1$$ hence this construction realizes every correlation $\geqslant\pi^2/6-2\approx-0.355$.

Did
  • 279,727