2017_III_13_ProbStats

2017_III_13_ProbStats

We have:

\displaystyle \mathrm V(x) = \mathrm E((X - x)^2) = \mathrm E(X^2 - 2x X + x^2) = \mathrm E(X^2) - 2 x \mathrm E(X) + x^2 = \mathrm E(X^2) - 2x \mu + x^2

Note that since \operatorname {var}(X) = \mathrm E(X^2) - (\mathrm E(X))^2 we have \mathrm E(X^2) = \sigma^2 + \mu^2, so:

\mathrm V(x) = \sigma^2 + \mu^2 - 2x \mu + x^2

We have:

Y = \sigma^2 + \mu^2 - 2X \mu + X^2

So:

\mathrm E(Y) = \sigma^2 + \mu^2 - 2 \mu \mathrm E(X) + \mathrm E(X^2) = \sigma^2 + \mu^2 - 2\mu^2 + \mu^2 + \sigma^2 = 2 \sigma^2

Note that for X uniformly distributed on [0,1] we have:

\displaystyle \mu = \int_0^1 x \mathrm dx = \frac 1 2

and:

\displaystyle \sigma^2 = \int_0^1 x^2 \mathrm dx - \left(\frac 1 2\right)^2 = \frac 1 3 - \frac 1 4 = \frac 1 {12}

so:

\displaystyle V(x) = \frac 1 {12} + \frac 1 4 - x + x^2 = x^2 - x + \frac 1 3 = \left(x - \frac 1 2\right)^2 + \frac 1 {12}

Note that for 0 \le X \le 1, we have \dfrac 1 {12} \le V(X) \le \dfrac 1 3.

We therefore have:

\begin{align*}\mathrm P(Y \le y) &= \mathrm P\left(\left(X - \frac 1 2\right)^2 + \frac 1 {12} \le y\right) \\ &= \mathrm P\left(\left(X - \frac 1 2\right)^2 \le y - \frac 1 {12}\right) \\ &= \mathrm P\left(\frac 1 2 - \sqrt {y - \frac 1 {12}} \le X \le \frac 1 2 + \sqrt {y - \frac 1 {12}}\right) \\ &= 2 \sqrt {y - \frac 1 {12}}\end{align*}

so:

\displaystyle f_Y(y) = \frac 1 {\sqrt {y - \frac 1 {12}}}

Then:

\begin{align*}\mathrm E(Y) &= \int_{\frac 1 {12}}^{\frac 1 3} \frac y {\sqrt {y - \frac 1 {12}}} \mathrm dy \\&= \int_{\frac 1 {12}}^{\frac 1 3} \frac {y - \frac 1 {12} + \frac 1 {12}} {\sqrt {y - \frac 1 {12}}} \mathrm dy \\&= \int_{\frac 1 {12}}^{\frac 1 3} \left(\sqrt {y - \frac 1 {12}} + \frac 1 {12 \sqrt {y - \frac 1 {12}}}\right) \mathrm dy \\&=\left[\frac {\left(y - \frac 1 {12}\right)^{3/2}} {\frac 3 2} + \frac 1 6 \sqrt {y - \frac 1 {12}}\right]_{\frac 1 {12}}^{\frac 1 3} \\ &= \frac 2 3 \times \left(\frac 1 3 - \frac 1 {12}\right)^{3/2} + \frac 1 6 \sqrt {\frac 1 3 - \frac 1 {12}} \\ &= \frac 2 3 \times \left(\frac 1 4\right)^{3/2} + \frac 1 6 \sqrt {\frac 1 4} \\ &= \frac 1 6\end{align*}

We have \sigma^2 = \dfrac 1 {12}, so we’re done.

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