Expected value of the product of two rvs

**Canadian0469** · January 26th 2009, 07:31 PM

Given two r.v.s Y1 and Y2 that have joint density fcn:

$f(y_1,y_2) = [1-\alpha[(1-2e^{-y_1})(1-2e^{-y_2})]]e^{-y_1-y_2}$ for 0 <= y_1, 0 <=y_2,
and 0 elsewhere

and $-1 \le \alpha \le 1$

I have to find $E(Y_1Y_2)$ .

I did this by doing:
$\int^\infty_0 y_1e^{-y_1}[\int^\infty_0y_2[1-\alpha[(1-2e^{-y_1})(1-2e^{-y_2})]]e^{-y_2}\,dy_2] \,dy_1$

But I end up at a different answer than the final answer in the back of my text. The answer given in the back is:
$E(Y_1Y_2) = 1 - \alpha/4$

**mr fantastic** · January 27th 2009, 02:52 AM

Originally Posted by Canadian0469

Given two r.v.s Y1 and Y2 that have joint density fcn:

$f(y_1,y_2) = [1-\alpha[(1-2e^{-y_1})(1-2e^{-y_2})]]e^{-y_1-y_2}$ for 0 <= y_1, 0 <=y_2,
and 0 elsewhere

and $-1 \le \alpha \le 1$

I have to find $E(Y_1Y_2)$ .

I did this by doing:
$\int^\infty_0 y_1e^{-y_1}[\int^\infty_0y_2[1-\alpha[(1-2e^{-y_1})(1-2e^{-y_2})]]e^{-y_2}\,dy_2] \,dy_1$

But I end up at a different answer than the final answer in the back of my text. The answer given in the back is:
$E(Y_1Y_2) = 1 - \alpha/4$

I agree with the answer $1 - \frac{\alpha}{4}$ .

Stage 1:

I get $\int^{+\infty}_0 y_2 \left[ 1 - \alpha (1-2e^{-y_1}) (1-2e^{-y_2}) \right] e^{-y_2} \,dy_2 = \frac{1}{2} \left( 2 e^{y_1} - \alpha \left[ e^{y_1} - 2\right] \right) \cdot e^{-y_1}$ .

Do you?

**Canadian0469** · January 27th 2009, 06:31 AM

No, I can't see where you got that.
I started off by expanding the inner expression (the integration with respect to $y_2$ , and then worked with
$<br /> \int^\infty_0 y_1e^{-y_1}[\int^\infty_0y_2e^{-y_2} - \alpha y_2e^{-y_2} + 2\alpha y_2 e^{-2y_2} + 2\alpha y_2 e^{-y_1-y_2} - 4\alpha y_2 e^{-y_1-2y_2} \,dy_2] \,dy_1<br />$

This was the only approach I could think of... Where did I go wrong?

**mr fantastic** · January 27th 2009, 11:26 AM

Originally Posted by Canadian0469

No, I can't see where you got that.
I started off by expanding the inner expression (the integration with respect to $y_2$ , and then worked with
$<br /> \int^\infty_0 y_1e^{-y_1}[\int^\infty_0y_2e^{-y_2} - \alpha y_2e^{-y_2} + 2\alpha y_2 e^{-2y_2} + 2\alpha y_2 e^{-y_1-y_2} - 4\alpha y_2 e^{-y_1-2y_2} \,dy_2] \,dy_1<br />$

This was the only approach I could think of... Where did I go wrong?

The answer I've given for the 'inner integral' is correct. You then use this result to do the 'outer integration'.

I have no idea where you're going wrong - to know that you will need to post all your working.

**Canadian0469** · January 27th 2009, 12:23 PM

Okay I have the same result for the inner integral now. I'll continue tonight and will post the result. Thanks for your support mr. fantastic!

**Canadian0469** · January 27th 2009, 06:34 PM

Okay, I finally got the correct answer. I really had to break it up the way you did to get it. It took less time than doing it the way I was doing it before (expanding the whole expression). I suppose I was making some sort of algebraic mistake before?

Thanks mr. fantastic, you are awesome!

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