# Thread: Expected value and variance

1. ## Expected value and variance

An electronic system has two components X and Y, and works if just one or both components are functioning. Their lifespan function is written as:

f(x,y) = 2^e−(x+2y), x > 0, y > 0, otherwise 0

I've calculted the marginal distributions to be:

g(x) = e^-x
h(y) = 2e^-2y

Next I'm supposed to figure out the expected lifespan for X and Y, plus the variance in lifespan for each. Would appreciate any help!

2. Originally Posted by gralla55
An electronic system has two components X and Y, and works if just one or both components are functioning. Their lifespan function is written as:

f(x,y) = 2^e−(x+2y), x > 0, y > 0, otherwise 0

I've calculted the marginal distributions to be:

g(x) = e^-x
h(y) = 2e^-2y

Next I'm supposed to figure out the expected lifespan for X and Y, plus the variance in lifespan for each. Would appreciate any help!
I haven't checked your answers (I assume you can integrate) but now that you have the pdf's for each random variable you just use them in the usual way.

3. Originally Posted by gralla55
An electronic system has two components X and Y, and works if just one or both components are functioning. Their lifespan function is written as:

f(x,y) = 2^e−(x+2y), x > 0, y > 0, otherwise 0

I've calculted the marginal distributions to be:

g(x) = e^-x
h(y) = 2e^-2y

Next I'm supposed to figure out the expected lifespan for X and Y, plus the variance in lifespan for each. Would appreciate any help!
Assuming your marginal distributions are correct.

$E(x) = \displaystyle\int^\infty_0 xe^{-x}\,dx = 1$
and $E(x^2) = \displaystyle\int^\infty_0 x^2e^{-x}\,dx = 2$
So, $Var(x) = 2 - 1^2 = 1$

Same procedure for y.

4. BY inspection these are two independent exponential rvs
The paramaters are 1 and 2.
Without integrating you should know the mean and variance of these two.