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This may be a dumb question, but I thought that to get the expected value, we were supposed to calculate

$$A = \int_0^1 \int_x^1 y 2(x+y) dydx$$

which gave me a different answer to

$$B = \int_0^1 \int_0^y y 2(x+y) dxdy$$

As far as I know the second expression is finding the marginal density $f_y$ first, then straightforwardly finding the expectation for that.

I keep finding that A = 5/12 while B = 3/4, but I am confident that A should also work, too.

So, it would really help if you could give me the following information.

1), Is there something wrong with my calculation?

2), If not, what DID I find with the expression A?

hyg17
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1 Answers1

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I get $\frac{3}{4}$ for A.

We are integrating $2xy+2y^2$ with respect to $y$. We get antiderivative $xy^2+\frac{2}{3}y^3$. Plugging in the bounds we get $x+\frac{2}{3}-\frac{5}{3}x^3$.

Integrating with respect to $x$ gives $\frac{x^2}{2}+\frac{2}{3}x-\frac{5}{12}x^4$.

Insert the bounds. We get $\frac{1}{2}+\frac{2}{3}-\frac{5}{12}$. Simplify.

For B we also get $\frac{3}{4}$. The calculation is shorter.

Remark: One can view the second way as finding the density function of $Y$ first. I prefer to think of the expectation as a double integral, evaluated in two ways as an iterated ntegral.

André Nicolas
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