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$X\sim U(0,1)$. Divide the interval [0,1] into k equal subintervals. Then $X_1$=the number of observtions on the first interval.

Define the new variable $Y_1=X_1/n$, where n is the number of observations, and $Y_1$ then is the proportion of the observation on the first interval.

What is the expected value, variance and standard deviation for $Y_1$?

From the definition of the expected value we have $E(Y)=\int\limits_\infty^\infty\mathrm{y}f(y)\mathrm{d}y$, if I could get help with this I guess i can do the variance and standard deviation. But how do i find $f(y$)?

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up vote 1 down vote accepted

I take it that you're drawing $n$ independent samples from $X\sim U(0,1)$. Then $X_1$ has a binomial distribution with parameters $n$ and $p=1/k$. The mean and variance of a binomial distribution are known, and $E[Y_1]=E[X_1]/n$ and $\operatorname{Var}(Y_1)=\operatorname{Var}(X_1)/n^2$.

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Thanks alot! Makes sense. Did I get this right? $\\$E(aX+b)=aE(X)+b; E($X_1/n$)=$\frac{E(X_1}{n}=\frac{n}{k*n}=\frac{1}{k}$ \*V(aX+b)=$a^2V(X); V(X_1/n=\frac{V(X_1)}{n^2}=n\frac{1}{k}\frac{k-1}{k}\frac{1}{n^2}=\frac{k-1}{k^2‌​n}$ \*$D(Y_1)=\sqrt{V(Y_1}=\frac{\sqrt{k-1}}{k\sqrt{n}}$ – Alexander Feb 3 '13 at 6:46
@Alexander: Yes, apart from the messed up formatting and parentheses, that all looks right to me. Sorry for the late reply. – joriki May 7 '13 at 9:00

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