Problem source

The random variable $X$ is
uniformly distributed on $[0,1]$. A new random variable
$Y$ is defined by the rule 
\[
Y=\begin{cases}
1/4 & \mbox{ if }X\leqslant1/4,\\
X & \mbox{ if }1/4\leqslant X\leqslant3/4\\
3/4 & \mbox{ if }X\geqslant3/4.
\end{cases}
\]
Find ${\mathrm E}(Y^{n})$ for all integers $n\geqslant 1$.
Show that ${\mathrm E}(Y)={\mathrm E}(X)$ and that
\[{\mathrm E}(X^{2})-{\mathrm E}(Y^{2})=\frac{1}{24}.\]
By using the fact that $4^{n}=(3+1)^{n}$, or otherwise,
show that ${\mathrm E}(X^{n}) > {\mathrm E}(Y^{n})$ for $n\geqslant 2$.
Suppose that $Y_{1}$, $Y_{2}$, \dots are independent random variables
each having the same distribution as $Y$.
Find, to a good approximation, $K$ such that
\[{\rm P}(Y_{1}+Y_{2}+\cdots+Y_{240000} < K)=3/4.\]

Solution source

\begin{align*}
&& \E[Y^n] &= \frac14 \cdot \frac1{4^n} + \frac14 \cdot \frac{3^n}{4^n} + \frac12 \int_{1/4}^{3/4}2 y^n \d y  \\
&&&= \frac{3^n+1}{4^{n+1}} + \left [ \frac{y^{n+1}}{n+1} \right]_{1/4}^{3/4} \\
&&&= \frac{3^n+1}{4^{n+1}} + \frac{3^{n+1}-1}{(n+1)4^{n+1}}
\end{align*}

\begin{align*}
&& \E[Y] &= \frac{3+1}{16} + \frac{9-1}{2 \cdot 16} \\
&&&= \frac{1}{4} + \frac{1}{4} = \frac12 = \E[X]
\end{align*}

\begin{align*}
&& \E[X^2] &= \int_0^1 x^2 \d x = \frac13 \\
&& \E[Y^2] &= \frac{9+1}{64} + \frac{27-1}{3 \cdot 64} = \frac{56}{3 \cdot 64} = \frac{7}{24} \\
\Rightarrow && \E[X^2] - \E[Y^2] &= \frac13 - \frac{7}{24} = \frac{1}{24}
\end{align*}

\begin{align*}
&& \E[X^n] &= \frac{1}{n+1} \\
&& \E[Y^n] &= \frac{1}{n+1} \frac{1}{4^{n+1}}\left ( (n+1)(3^n+1)+3^{n+1}-1 \right) \\
&&&= \frac{1}{n+1} \frac{1}{4^{n+1}}\left ( 3^{n+1} + (n+1)3^n +n \right) \\
\\
&& (3+1)^{n+1} &= 3^{n+1} + (n+1)3^n + \cdots + (n+1) \cdot 3 + 1 \\
&&&> 3^{n+1} + (n+1)3^n + n + 1
\end{align*}
if $n \geq 2$

Notice that by the central limit theorem:

\begin{align*}
&&\frac{1}{240\,000} \sum_{i=1}^{240\,000} Y_i &\sim N \left ( \frac12, \frac{1}{24 \cdot 240\,000}\right) \\
\Rightarrow && \mathbb{P}\left (\frac{\frac{1}{240\,000} \sum_{i=1}^{240\,000} Y_i  - \frac12}{\frac1{24} \frac{1}{100}} \leq \frac23 \right) &\approx 0.75 \\

\Rightarrow && \mathbb{P} \left ( \sum_i Y_i \leq 240\,000 \cdot \left ( \frac2{3} \frac1{2400}+\frac12 \right) \right ) & \approx 0.75 \\
\Rightarrow && K &= 120\,000 + 66 \\
&&&\approx 120\,066
\end{align*}