I need to calculate the Covariance of two random variables $X$ and $Y$ with exponential distributions.
The probability density functions are $f(x)=\lambda \cdot e^{-\lambda x}$ for $x \in [0, \infty)$ and $f(y)=\lambda \cdot e^{-\lambda (y-b)}$ where $b \in \mathbb{R^+}$ for $y \in [b, \infty)$
My lecture states that $ Cov(X,Y)= E[XY] - E[X]E[Y] $.
I know that $ E[X]= \int xf(x)dx $
Here is the Question: How do I calculate $E[XY]$ ? I would guess I need to calculate a joint density function $f_{XY} (x,y)$ to then calculate the expected value $ E[XY] = \int_b^\infty \int_0^\infty xyf_{XY}(x,y)dxdy$
However I don´t know how to calculate the joint density function. Nor do I know how to determine if the Variables are independent, if they were, the covariance would of course be zero.
EDIT: I should add: X is the random variable for time to failure of a system. Y is the random variable for time to failure + set amount of time b. I am not sure wether it was correct to construct the pdf for Y the way I did. Now I feel like, intuitively, X and Y should be dependent on each other and strongly correlate.
- Edited values of 5 to be b to keep the question more general( 5 is the value of b given in the exercise, the value $\lambda = 1/1700$ is also given)
- For the full exercise, see the pictureExcercise about exponential distributions
After consulting with the lecturer, the trick seems to be realizing that:
$$ Y = X+b$$
which can be used to rewrite
$$ Cov(X,Y) = E[XY] - E[X]E[Y] = E[X(X+b)] - E[X]E[X+b]$$
to receive
$$Cov(X,Y) = E[X^2]+bE[X] - E[X](E[X]+E[b]) = E[X^2]+bE[X] - E[X]^2-bE[X]= Var[X]$$
with $ E[b] = b$ , because $b$ is a constant
and $ E[X^2] - E[X]^2 = Var[X]$
All this leads to the realization that the coefficient of correlation is 1
$$ r_{xy} = {Cov(X,Y) \over \sqrt{Var[X]Var[Y]}} = 1$$
since $Y = X+b$, meaning they are perfectly correlated.
All this makes it possible to circumvent calculating a joint density/distribution function and from there calculating the integrals for the different expectation values.