The Gershgorin circle theorem states that every eigenvalue of $A$ lies in at least one discFor a nonzero complex eigenvector $v$, choose an index $i$ with $|v_i|={\max }_j|v_j|>0$. The $i$th row of $Av=\lambda v$ givesDividing by $|v_i|$ proves membership in $D_i$. This is valid for real matrices even when the eigenvalues are complex.

Let $F_{n-1}$ be the cumulative distribution function of the Student t-distribution. The two-sided p-value is $2[1-F_{n-1}(|t_0|)]$. If it is less than $\alpha$, strict monotonicity gives $|t_0|>t_{n-1,1-\alpha /2}$, henceThe magnitude of the interval's centre is larger than its half-width, so zero lies outside the confidence interval. This is the duality between confidence intervals and hypothesis tests.

Under $H_0:\mu =0$, the statistic $T=\sqrt{n}\overline{X}/S$ has Student t-distribution with $n-1$ degrees of freedom. Reject in either tail, splitting the null error probability equally. Thus the critical region isThe distribution is continuous, so the boundary convention does not change the size $\alpha$.

Let $t_{\nu ,\beta }$ denote a quantile of the Student t-distribution with $\nu =n-1$ degrees of freedom. Symmetry gives $\mathbb{P}(|Y|\le t_{n-1,1-\alpha /2})=1-\alpha$. Solving for $\mu$ gives the exact central confidence interval

For a normal sample with unknown mean and variance, the sample mean and sample variance are independent, with $\sqrt{n}(\overline{X}-\mu )/\sigma \sim N(0,1)$ and $(n-1)S^2/{\sigma }^2\sim {\chi }_{n-1}^2$. Therefore the ratio defining the Studentized normal sample mean hasusing $S^2=(n-1{)}^{-1}{\sum }_i(X_i-\overline{X}{)}^2$.

The total $S$ under $H_0$ has Poisson distribution with mean $12$, and the observed total is $16$. Its one-sided p-value is thereforeThe supplied ${\varphi }_{12}(17)\approx 0.1$ also shows that the observed count does not enter the approximate 10% upper-tail rejection region. Thus there is insufficient evidence to reject $\mu =1$ at either the 10% or 5% level. The question specifies no testing level, so the p-value gives the general conclusion: rejection is justified only at levels at least about $0.1556$.

The printed ${\varphi }_1(16)\approx 0$ concerns a Poisson variable with mean one, not the twelve-month total. It is therefore not the null tail for this data; substituting it would incorrectly treat the yearly count as one month's count.

For $S={\sum }_i{X}_i$, the ratio of the two sample likelihood functions isFor $0<{\mu }_0<{\mu }_1$ it is strictly increasing in $S$. The Neyman-Pearson lemma therefore selects an upper tail:Under the null, $S\sim \mathrm{Poisson}(n{\mu }_0)$, so choose $k$ with this upper-tail probability at most $\alpha$. If exact size $\alpha$ is required, randomize at the adjacent boundary value as needed. The rejection rule is the same for every ${\mu }_1>{\mu }_0$, giving a uniformly most powerful test for the one-sided alternative.

The Neyman-Pearson lemma concerns testing a simple null density $f_0$ against a simple alternative $f_1$. A test that rejects where $f_1/f_0>c$, accepts where the ratio is below $c$, and if necessary randomizes on the boundary to have null rejection probability $\alpha$, is a most powerful test among tests of size at most $\alpha$. Thus a likelihood-ratio threshold maximizes power at the specified simple alternative. The boundary randomization is relevant for discrete distributions.

Since the true mean is known to be zero, the chi-squared distribution gives $Q/\theta \sim {\chi }_n^2$. Let $q_{\beta }$ be the $\beta$ quantile of this law. The probabilityrearranges, using positivity, to the exact confidence intervalThere are $n$ degrees of freedom, because no unknown mean was fitted.

Put $Q={\sum }_i{X}_i^2$. For $\theta >0$, the log-likelihood is $\ell (\theta )=\text{constant}-(n/2)\log \theta -Q/(2\theta )$. Its derivative is $-n/(2\theta )+Q/(2{\theta }^2)$, changing from positive to negative at $Q/n$. ThusThe known mean is zero, so no centring at the sample mean is needed. Taking the expected negative second derivative gives $I_n(\theta )=n/(2{\theta }^2)$. The large-sample variance is consequently $2{\theta }^2/n$. With $z_{1-\alpha /2}$ the indicated standard-normal quantile, a Wald confidence interval isThe interval may be restricted to the parameter space $\theta >0$. The parameter is the variance itself, so the standard error is proportional to $\hat{\theta }$, not to its square root.

The marginal probability density function is $f_Y(y)={\int }_{\mathbb{R}}{f}_{X,Y}(x,y)dx$. For $y$ with $f_Y(y)>0$, the conditional density isIf the conditional first absolute moment is finite, the conditional expectation isConditioning here uses a density, since the individual event $Y=y$ normally has probability zero; a conditional law is specified only up to marginal-null values of $y$.

With probability $2/3$, the chain starts at $+1$ and $T=1$. Conditional on starting at $-1$, it takes a geometric number $G$ of transitions to first reach $+1$, with success probability $1-p$, so $T=1+G$. The probability generating function is thereforeThe parameter $q$ is absent because the process stops at its first visit to $+1$. At $s=1$ the expression equals one, confirming eventual hitting with probability one.

The finite irreducible aperiodic Markov chain converges to its stationary distribution, regardless of its initial law. Therefore $\mathbb{E}[X_j]\to {\pi }_{+}-{\pi }_{-}$. Taking the arithmetic means of this convergent sequence givesThis uses linearity of expectation and convergence of Cesàro means; it does not require the variables $X_j$ to be independent.