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The question states: "Let Y be a continuous random variable which is uniform over (0, 1). Find the expected value and variance of Y. Then prove a general formula for the rth moment about the origin and the mean."

So I know the expected value/mean is $E[Y] = 0.5 = \mu$ and the equation of variance to be: $$ Var(Y) = E[(Y-\mu)^2] = \int_0^1(y-\mu)^2 \cdot f_Y(y)\space dy $$ But I don't know what $f_Y(y)$ is or what the value of $E[Y^2]$ would be (Would it also just be 0.5?)

And the r$^{th}$ moment of Y about the origin is: $$\mu_T = E[W^T] $$ I really don't understand how I could write a general formula for this, other than it always being equal to 0.5. Is this correct at all? I'm really not very confident in my answers.

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$f_Y(f) = \frac{1}{b-a}$ for a uniform random variable over the range of a to b, 0 otherwise. In your case a = 0 and b = 1.

$E[g(Y)] = \int_0^1 g(y) f_Y(y) dy$ so $E[Y^2] = \int_0^1 y^2 f_Y(y) dy$

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If $Y$ is a random variable with density of $f_Y(y)$ and $\psi$ is nice enough,

$$E(\psi(Y))=\int\psi(y)f_Y(y)dy$$

So in your case, let $\psi(y)=y^r$ to get:

$$E(Y^r)=\int y^rf_Y(y)dy$$

The density for a uniform $(a,b)$ random variable is simply:

$$f(y)=\begin{cases}\frac{1}{b-a}&\text{ if } y\in(a,b)\\0&\text{ else}\end{cases}$$

So:

$$E(Y^r)=\int_0^1y^rdy$$

I hope you can compute this.

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