Problem source

Two coins $A$ and $B$ are tossed together. $A$ has  
probability $p$ of showing a head, and $B$ has probability $2p$, independent of $A$,
of showing a head,
where $0 < p < \frac12$.
The random variable $X$ takes the value 1 if $A$ 
shows a head and it takes the value $0$  if $A$ shows a tail. 
The random variable $Y$ takes the value 1 if $B$ 
shows a head and it takes the value $0$  if $B$ shows a tail.
The random variable $T$ is defined by
\[
T= \lambda X + {\textstyle\frac12} (1-\lambda)Y.
\]
Show that $\E(T)=p$ and find an expression for $\var(T)$ in terms of $p$ and $\lambda$.
Show that as $\lambda$ varies, the minimum of $\var(T)$ occurs when 
\[
\lambda =\frac{1-2p}{3-4p}\;.
\]
The two coins are tossed $n$ times, where $n>30$, and $\overline{T}$ is the mean value of $T$.
Let $b$ be a fixed positive number. Show that the maximum value of 
$\P\big(\vert \overline{T}-p\vert < b\big)$ as $\lambda$ varies is approximately $2\Phi(b/s)-1$,
 where $\Phi$ is the cumulative distribution function of a standard normal variate and 
\[
s^2= \frac{p(1-p)(1-2p)}{(3-4p)n}\;.
\]

Solution source

\begin{align*}
&& \E[T] &= \E[\lambda X + \tfrac12(1-\lambda)Y] \\
&&&= \lambda \E[X] + \tfrac12(1-\lambda) \E[Y] \\
&&&= \lambda p + \tfrac12 (1-\lambda) 2p \\
&&&= p \\
\\
&& \var[T] &= \var[\lambda X + \tfrac12(1-\lambda)Y] \\
&&&= \lambda^2 \var[X] + \tfrac14(1-\lambda)^2 \var[Y] \\
&&&= \lambda^2 p(1-p) + \tfrac14(1-\lambda)^22p(1-2p) \\
&&&= p(\lambda^2 + \tfrac12(1-\lambda)^2) -p^2(\lambda^2+(1-\lambda)^2)\\
&&&= p(\tfrac32\lambda^2 - \lambda + \tfrac12) -p^2(2\lambda^2 -2\lambda + 2)
\end{align*}

Differentiating $\var[T]$ with respect to $\lambda$, and noting it is a quadratic with positive leading coefficient, we get

\begin{align*}
&& \frac{\d \var[T]}{\d \lambda} &= p(2\lambda -(1-\lambda)) - p^2(2 \lambda -2(1-\lambda)) \\
&&&= p(3\lambda - 1)-p^2(4\lambda - 2) \\
\Rightarrow && \lambda(4p-3) &= 2p-1 \\
\Rightarrow && \lambda &= \frac{1-2p}{3-4p}
\end{align*}

By the central limit theorem $\overline{T} \sim N(p, \frac{\sigma^2}{n})$ in particular, $\mathbb{P}(|\overline{T} - p| < b) = \mathbb{P}(\left \lvert |\frac{\overline{T}-p}{\frac{\sigma}{\sqrt{n}}} \right \lvert <  \frac{b}{\frac{\sigma}{\sqrt{n}}}) = \mathbb{P}(|Z| < \frac{b\sqrt{n}}{\sigma}) = 2\Phi(b/s) - 1$ where $s = \frac{\sigma}{\sqrt{n}}$ so

\begin{align*}
&& s^2 &= \frac1n \sigma^2 \\
&&&= \frac1n \left ( \left (\left ( \frac{1-2p}{3-4p} \right)^2 + \tfrac12 \left (1-\frac{1-2p}{3-4p} \right)^2 \right)p - \left ( \left ( \frac{1-2p}{3-4p} \right)^2 +  \left (1-\frac{1-2p}{3-4p} \right)^2\right)p^2 \right) \\
&&&= \frac1n \left ( \left (\left ( \frac{1-2p}{3-4p} \right)^2 + \tfrac12 \left (\frac{2-2p}{3-4p} \right)^2 \right)p - \left ( \left ( \frac{1-2p}{3-4p} \right)^2 +  \left (\frac{2-2p}{3-4p} \right)^2\right)p^2 \right) \\
&&&= \frac{p}{n(3-4p)^2} \left ( (1 -4p + 4p^2 + 2-4p+2p^2) - (1-4p+4p^2+4-8p+4p^2)p \right) \\
&&&= \frac{p}{n(3-4p)^2} \left (3-13p+18p^2-8p^3   \right) \\
&&&= \frac{p}{n(3-4p)^2} (3-4p)(1-2p)(1-p) \\
&&&= \frac{p(1-p)(1-2p)}{(3-4p)n}

\end{align*}