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Let $X$ and $Y$ be random independent variables within the limits $[0, 1]$ with the following density functions:

$f_X(x) = 0.16x + 0.92$ such that $x$ is within the parameters $[0, 1]$ and $f_Y(y) = 1.41y^2 + 0.53$ such that $y$ is within the parameters $[0, 1]$.

how do I calculate $E[X^²Y^5]$?

I used the the formula $E[X] = \frac{1}{λ}$, solved for $λ$, then substituted it into the equation $E[X^n] = \frac{n!}{\lambda^n}$. I calculated this for $E[X^2]$ and $E[Y^5]$ separately, then multiplied them.

I got the wrong answer but I don't know how else to solve it.

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  • $\begingroup$ You can separate those like that if they're independent, so you're safe. $\endgroup$ Nov 20, 2013 at 1:58

2 Answers 2

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Just to further clarify Potato's answer, as far as the exponents go:

$E(X^2Y^5)=E(X^2)E(Y^5)$

$E(X^2) = E(g(X))$

$E(Y^5) = E(h(X))$

And from there we can move on to integrating. You should read about the law of the unconscious statistician.

Just as an example, integrating $E(X^2)$ would look like:

$\int g(x) f(x) dx,$

Then substituting in $g(x) = x^2$ and the density function, and adding the limits, we get:

$\int_0^1 x^2 (0.16x+0.92) dx$

Do the same for E(Y), and you should be good to go.

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Yes, functions of independent variables are independent, so $X^2$ and $Y^5$ are independent. Then $E(X^2Y^5)=E(X^2)E(Y^5)$ and your method works.

I'm not sure why you're trying to use that particular formula. To find the expectation, you just integrate $xf_x(x)$ across all of $\mathbb R$. You only need to use the integral over $[0,1]$, since $f_x$ vanishes outside it. Then do the same for $f_y$. For more information, see here.

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