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\textit{\underline{Solution 1.4}: we have by definition
$$
        \mathbb{E} (T_n - \mu)^2 = \mathbb{E} \left[(\overline{X}_n - \mu)^2 1_{|\overline{X}_n| > n^{-1/4}} \right] + \mu^2 \mathbb{P} \left(|\overline{X}_n| \leq n^{-1/4} \right).
$$
At $\mu = 0$, it simplifies to
$$
	\mathbb{E} T_n^2 = \mathbb{E} \left[\overline{X}_n^2 1_{|\overline{X}_n| > n^{-1/4}} \right].
$$
The boundedness of $\overline{X}_n$ and Cauchy-Schwarz's inequality lead to}
$$ 
        \mathbb{E} T_n^2 = \mathbb{E} \left[\overline{X}_n^2 1_{|\overline{X}_n| > n^{-1/4}} \right] \leq \sqrt{\mathbb{E} \left[\overline{X}_n^4\right] \mathbb{P}\left(|\overline{X}_n| > n^{-1/4}\right)} \leq 2b^2 \exp\left(-\frac{\sqrt{n}}{4b^2}\right). \qquad \square 
$$\\

5. Show that for any fixed $\mu \neq 0$, $R(T_n, \mu)$ converges to zero at rate $O(1/n)$.\\

\textit{\underline{Solution}: thanks to the triangle inequality $|\overline{X}_n -\mu| \geq |\mu| - |\overline{X}_n|$, we deduce
$$
        \mathbb{P} \left( |\overline{X}_n| \leq |\mu| - n^{-1/4}\right) \leq \mathbb{P}\left(|\overline{X}_n - \mu| \geq n^{-1/4}\right) \leq 2\exp\left(-\frac{\sqrt{n}}{2b^2}\right). 
$$
When $\mu \neq 0$, as soon as $|\mu| > 2n^{-1/4}$, we obtain
$$
          \mathbb{P} \left(|\overline{X}_n| \leq n^{-1/4} \right) \leq \mathbb{P} \left( |\overline{X}_n| \leq |\mu| - n^{-1/4}\right) \leq 2\exp\left(-\frac{\sqrt{n}}{2b^2}\right). 
$$
Again, writing out the expression for the quadratic risk 
\begin{align*}
        \mathbb{E} (T_n - \mu)^2 & = \mathbb{E} \left[(\overline{X}_n - \mu)^2 1_{|\overline{X}_n| > n^{-1/4}} \right] + \mu^2 \mathbb{P} \left(|\overline{X}_n| \leq n^{-1/4} \right)\\ 
				& = \mathbb{E} \left[(\overline{X}_n - \mu)^2 \right] - \mathbb{E} \left[(\overline{X}_n - \mu)^2 1_{|\overline{X}_n| \leq n^{-1/4}} \right]  + \mu^2 \mathbb{P} \left(|\overline{X}_n| \leq n^{-1/4} \right),
\end{align*}
we find that the second and third term decrease much faster than the first one. Hence the conclusion. $\square$\\
}


\textit{\underline{Solution 1.5}: we lower bound the MSE
$$
        \mathbb{E} (T_n - \mu_n)^2 = \mathbb{E} \left[(\overline{X}_n - \mu_n)^2 1_{|\overline{X}_n| > n^{-1/4}} \right] + \mu_n^2 \mathbb{P} \left(|\overline{X}_n| \leq n^{-1/4} \right) \geq \mu_n^2 \mathbb{P} \left(|\overline{X}_n| \leq n^{-1/4} \right)
$$
as the first term decreases at least as fast as $O(1/n)$. Since we have by the concentration inequality
$$
        \mathbb{P} \left(|\overline{X}_n| \leq n^{-1/4} \right) \geq \mathbb{P} \left(|\overline{X}_n - \mu_n| \leq \frac{1}{2}n^{-1/4} \right) \geq 1 - 2 \exp\left( -\frac{\sqrt{n}}{8b^2}\right), 
$$
it follows immediately that
$$
        \lim_{n\to \infty}\mu_n^2 \mathbb{P} \left(|\overline{X}_n| \leq n^{-1/4} \right) = \frac{1}{4\sqrt{n}}
$$
hence the result. $\square$
}\\

\textit{\underline{Solution 2.2}: Let $\eta$ denote the LRT's threshold and we want it to satisfy 
$$
        \mathbb{P}\left( (\boldsymbol{b} - \boldsymbol{a})^T \Sigma^{-1} Y > \eta | X = 0 \right) = \alpha.
$$
Note that under $X = 0$, the test statistic is
$$
        (\boldsymbol{b} - \boldsymbol{a})^T \Sigma^{-1} Y  =  (\boldsymbol{b} - \boldsymbol{a})^T \Sigma^{-1} (\boldsymbol{a} + N) \sim \mathcal{N}\left((\boldsymbol{b} - \boldsymbol{a})^T \Sigma^{-1} \boldsymbol{a}, (\boldsymbol{b} - \boldsymbol{a})^T \Sigma^{-1} (\boldsymbol{b} - \boldsymbol{a}) \right).
$$
Therefore, let $\gamma := \sqrt{(\boldsymbol{b} - \boldsymbol{a})^T \Sigma^{-1} (\boldsymbol{b} - \boldsymbol{a})}$. The rejection region is 
\begin{align*}
\left\{\boldsymbol{y} \in \mathbb{R}^d, \;\; (\boldsymbol{b} - \boldsymbol{a})^T \Sigma^{-1} (\boldsymbol{y} - \boldsymbol{a}) \geq \gamma F^{-1}(1-\alpha)\right\}
\end{align*}
where $F^{-1}(\cdot)$ denotes the quantile of the standard normal distribution. $\square$
}\\

\textit{\underline{Solution 2.3}: again, let us define the rejection region 
$$
	R = \{ \boldsymbol{y} \in \mathbb{R}^d, \quad (\boldsymbol{b} - \boldsymbol{a})^T \Sigma^{-1} Y > \eta  \}.
$$
The type I error associated with the threshold $\eta$ is
$$
        \mathbb{P}\left( (\boldsymbol{b} - \boldsymbol{a})^T \Sigma^{-1} (Y - \boldsymbol{a})  > \eta - (\boldsymbol{b} - \boldsymbol{a})^T \Sigma^{-1}\boldsymbol{a} | X = 0 \right)
$$
and its type II error is 
$$
        \mathbb{P}\left( (\boldsymbol{b} - \boldsymbol{a})^T \Sigma^{-1} (Y - \boldsymbol{b})  \leq \eta - (\boldsymbol{b} - \boldsymbol{a})^T \Sigma^{-1}\boldsymbol{b} | X = 1 \right)
$$
It follows from the requirement that the two errors are equal
$$
	\eta - (\boldsymbol{b} - \boldsymbol{a})^T \Sigma^{-1}\boldsymbol{b} = -\eta + (\boldsymbol{b} - \boldsymbol{a})^T \Sigma^{-1}\boldsymbol{a}
$$
that is
$$
	\eta = \frac{1}{2} (\boldsymbol{b} - \boldsymbol{a})^T \Sigma^{-1}(\boldsymbol{b} +  \boldsymbol{a}).  \quad \square
$$
}\\

\textit{\underline{Solution 2.4}: it suffices to look at the type I error, which can be written as
$$
\mathbb{P}\left( (\boldsymbol{b} - \boldsymbol{a})^T \Sigma^{-1} N  > \frac{1}{2} (\boldsymbol{b} - \boldsymbol{a})^T \Sigma^{-1}(\boldsymbol{b} -  \boldsymbol{a}) \right) = 1 - F\left(\frac{\gamma}{2}\right).
$$
Hence, the stronger the SNR, the smaller the error. $\square$
}

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