Hypothesis test between two normal distributions Let $T_1,T_2,\ldots ,T_ n$ be i.i.d. observations, each drawn from a common normal distribution with mean zero. With probability $1/2$ this normal distribution has variance $1$, and with probability $1/2$ it has variance $4$. Based on the observed values $t_1,t_2,\ldots ,t_ n$, we use the MAP rule to decide whether the normal distribution from which they were drawn has variance $1$ or variance $4$. The MAP rule decides that the underlying normal distribution has variance $1$ if and only if$$\left| c_1 \sum _{i=1}^{n} t_ i^2 + c_2 \sum _{i=1}^{n} t_ i \right| < 1.$$
Find the values of $c_1\geq 0$ and $c_2\geq 0$ such that this is true. Express your answer in terms of $n$.
I'm finding it conceptually difficult to understand the nature of the posterior distribution of the variance of a normal distribution, given the observed $t_ i$.I'd greatly appreciate it if someone would please provide some helpful hint. Thanks.
 A: Using Bayes' Rule, the posterior distribution of the variance is given by the product of the likelihood and the prior distribution of the variance divided by the marginal likelihood
$$p(\sigma^2|t_1,t_2,\cdots,t_n)=\frac{p(t_1,t_2,\cdots,t_n|\sigma^2)p(\sigma^2)}{ p(t_1,t_2,\cdots,t_n)}=\frac{p(t_1,t_2,\cdots,t_n|\sigma^2)p(\sigma^2)}{\int p(t_1,t_2,\cdots,t_n|\sigma^2)p(\sigma^2)d(\sigma^2)} $$
As the observed values $t_i$ are i.i.d. (with zero mean) the likelihood function $p(t_1,t_2,...,t_n|\sigma^2)$  is given by
$$p(t_1,t_2,...,t_n|\sigma^2)=\prod_{i=1}^n\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{t_i^2}{2\sigma^2}\right)=\frac{1}{\sigma^n(2\pi)^{n/2}}\exp\left(-\frac{1}{2\sigma^2}\sum_{i=1}^nt_i^2\right)$$
while the prior on the variance is simply the following probability mass function
$$p(\sigma^2=1)=\frac{1}{2},\ p(\sigma^2=4)=\frac{1}{2}$$ 
Given the discrete nature of the prior of the variance, the marginal likelihood simplifies to
$$\begin{align}p(t_1,t_2,\cdots,t_n)&=p(t_1,t_2,\cdots,t_n|\sigma^2=1)p(\sigma^2=1)+p(t_1,t_2,\cdots,t_n|\sigma^2=4)p(\sigma^2=4)\\&=\frac{1}{2}\left[\frac{1}{(2\pi)^{n/2}}\exp\left(-\frac{1}{2}\sum_{i=1}^nt_i^2\right)+\frac{1}{2^n(2\pi)^{n/2}}\exp\left(-\frac{1}{8}\sum_{i=1}^nt_i^2\right)\right]\end{align}$$
In order to use the MAP rule to determine whether the samples were drawn from a distribution where the variance was $1$, we need to satisfy the following inequality (have a look at Equations $3.4$ and $3.5$ in here):-
$$\frac{p(t_1,t_2\cdots,t_n|\sigma^2=4)p(\sigma^2=4)}{p(t_1,t_2,\cdots,t_n)}<\frac{p(t_1,t_2\cdots,t_n|\sigma^2=1)p(\sigma^2=1)}{p(t_1,t_2,\cdots,t_n)}$$
which (given the prior of the variance) simplifies to
$$p(t_1,t_2\cdots,t_n|\sigma^2=4)<p(t_1,t_2\cdots,t_n|\sigma^2=1)$$
which is the condition that
$$\frac{1}{2^n(2\pi)^{n/2}}\exp\left(-\frac{1}{8}\sum_{i=1}^nt_i^2\right)<\frac{1}{(2\pi)^{n/2}}\exp\left(-\frac{1}{2}\sum_{i=1}^nt_i^2\right)$$
Taking the logarithm of both sides and simplifying results in the following inequality
$$\frac{3}{8n\log 2}\sum_{i=1}^nt_i^2<1$$
Thus, we have $c_1=\frac{3}{8n\log 2}$ and $c_2=0$.
