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.


$f_Y(f) = \frac{1}{b-a}$ 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 $E[Y^2] = \int_0^1 y^2 f_Y(y) dy$


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


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:

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



I hope you can compute this.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.