This note is still being edited: the text may contain typos, inaccuracies, and locally unpolished passages. If you find something, please let the author know (contact information on the course page).
In previous chapters we studied point estimators of parameters. Of course, in most cases the probability that the true value of the parameter coincides exactly with its estimator is zero. For this reason it is more convenient to consider, as an estimate, an entire interval of values of the unknown parameter. At the same time, the size of this interval should of course be as small as possible, while the probability of the true value of the parameter falling into it should, on the contrary, be as large as possible. In this chapter we discuss such intervals.
Definition and Methods of Constructing Confidence Intervals
Let \(X-\) be an observation with unknown distribution \(\mathbb {P} \in \mathcal{P}= \left\{ \mathbb {P}_\theta , \; \theta \in \Theta \right\}\).
Definition 1 A pair of statistics \(\left(T_{1}(X), T_{2}(X)\right)\) (where \(X\) is a sample) is called a confidence interval of confidence level \(\gamma\) for the parameter \(\theta\) if for every \(\theta \in \Theta \subset \mathbb {R}\) we have \[
\mathbb {P}_{\theta }\left(T_{1}(X)<\theta <T_{2}(X)\right) \geq \gamma
\] A confidence interval is called exact if for every \(\theta \in \Theta\) equality holds instead of the last inequality.
Constructing [Inexact] Confidence Intervals Using Chebyshevβs Inequality
To find a confidence interval one can, for example, use Markovβs inequality or Chebyshevβs inequality.
Let \(X_{1}, \ldots X_{n}\) be a sample from the Poisson distribution with parameter \(\lambda >0\). Construct a confidence interval for \(\lambda\) of confidence level \(\gamma\).
By Chebyshevβs inequality,
\[
\mathbb {P}_{\lambda }\left(\left|\sum X_{i}-\lambda n\right| \geq \varepsilon n\right) < \frac{\lambda }{\varepsilon ^{2} n}
\]
Hence,
\[
\mathbb {P}_{\lambda }\left(\lambda -\varepsilon < \overline{X} < \lambda +\varepsilon \right) \geq 1 - \frac{\lambda }{\varepsilon ^2 n} = \gamma
\]
From the last equality we find \(\varepsilon =\sqrt{\lambda /((1-\gamma ) n)}\). Denote \(c := \frac{1}{(1-\gamma )n}\), then \(\varepsilon = \sqrt{c\lambda }\), and it remains to solve the inequalities \(\overline{X}<\lambda +\sqrt{c\lambda }\) and \(\lambda - \sqrt{c\lambda } < \overline{X}\) for \(\lambda\).
Let us start with the first inequality. Set \(u := \sqrt{\lambda } \geq 0\); then it can be rewritten as
\[
u^2 + \sqrt{c}\, u - \overline{X} > 0.
\]
Complete the square:
\[
\left(u + \frac{\sqrt{c}}{2}\right)^2 - \frac{c}{4} - \overline{X}> 0 \quad \iff \quad \left(u + \frac{\sqrt{c}}{2}\right)^2 > \overline{X} + \frac{c}{4}.
\]
Since \(u + \frac{\sqrt c}{2} > 0\), we may take the square root of both sides:
\[
u > \sqrt{\overline{X}+\frac{c}{4}} - \frac{\sqrt{c}}{2}, \qquad \text{i.e.} \qquad \lambda = u^2 > \left(\sqrt{\overline{X}+\frac c4} - \frac{\sqrt c}{2}\right)^2.
\]
The second inequality (\(\lambda - \sqrt{c\lambda } < \overline{X}\)) is solved similarly with the same substitution \(u = \sqrt\lambda\) (only now the completed square is \(\left(u - \frac{\sqrt c}{2}\right)^2 < \overline{X} + \frac c4\)), giving \(\lambda < \left(\sqrt{\overline{X}+\frac c4} + \frac{\sqrt c}{2}\right)^2\). Finally, substituting \(c = \frac{1}{(1-\gamma )n}\) (so that \(\frac{\sqrt c}{2} = \sqrt{\frac{1}{4(1-\gamma )n}}\)), we obtain
\[
\mathbb {P}_{\lambda }\left(\left(\sqrt{\overline{X}+\frac{1}{4(1-\gamma ) n}}-\sqrt{\frac{1}{4(1-\gamma ) n}}\right)^2< \lambda < \left(\sqrt{\overline{X}+\frac{1}{4(1-\gamma ) n}}+\sqrt{\frac{1}{4(1-\gamma ) n}}\right)^2\right) \geq \gamma
\]
A more precise (though not exact in the strict sense) interval can be obtained by using the fact that \(\sum_{i=1}^n X_i \sim \operatorname {Pois}\left(n\lambda \right)\), and taking quantiles of this distribution instead of Chebyshevβs inequality β see the asymptotic version of this example at the end of the chapter.
Constructing Exact Confidence Intervals: the Central Statistic Method
Chebyshevβs inequality, as a rule, gives a very crude bound on the probability of deviation from the mean. The diameter of such an interval usually turns out to be excessively large. Let us try to construct an exact interval.
Let \(X_{1}, \ldots , X_{n}\) be a sample from the distribution \(\mathscr {N}\left(\theta , \sigma^2\right)\) with known \(\sigma^2\). Construct an exact confidence interval for \(\theta\) of confidence level \(\gamma\).
Note that \(X_1 + \ldots + X_n \sim \mathscr {N}\left(n\theta , n\sigma^2\right)\) and \(\overline{X} \sim \mathscr {N}\left(\theta , \frac{\sigma^2}{n}\right)\). Hence,
\[
\overline{X} - \theta \sim \mathscr {N}\left(0, \frac{\sigma ^2}{n}\right), \qquad G(X, \theta ) := \frac{\overline{X} - \theta }{\sigma /\sqrt{n}} \sim \mathscr {N}\left(0, 1\right).
\]
Notice that the distribution of the random variable \(G(X, \theta )\) does not depend on the unknown parameter \(\theta\). Such a random variable is called a central statistic. Let \(u_{\frac{1-\gamma }{2}}, u_{\frac{1 + \gamma }{2}}\) be the quantiles of levels \(\frac{1-\gamma }{2}, \frac{1+\gamma }{2}\) of the distribution \(\mathscr {N}\left(0, 1\right)\). By the definition of quantiles,
\[
\gamma = \frac{1 + \gamma }{2} - \frac{1 - \gamma }{2} = \mathbb {P}\left(u_{\frac{1-\gamma }{2}} < \frac{\overline{X} - \theta }{\sigma /\sqrt{n}} < u_{\frac{1+\gamma }{2}}\right).
\]
Let us solve the inequality under the probability for \(\theta\): multiplying by \(\sigma /\sqrt{n}\), we get
\[
\frac{\sigma }{\sqrt{n}}u_{\frac{1-\gamma }{2}} < \overline{X} - \theta < \frac{\sigma }{\sqrt{n}}u_{\frac{1+\gamma }{2}},
\]
whence (subtracting \(\overline{X}\) and changing sign, i.e. reversing the inequality)
\[
\overline{X} - \frac{1}{\sqrt{n}}u_{\frac{1+\gamma }{2}}\sigma < \theta < \overline{X} - \frac{1}{\sqrt{n}}u_{\frac{1-\gamma }{2}}\sigma .
\]
Hence,
\[
\gamma = \mathbb {P}\left(\theta \in \left( \overline{X} - \frac{1}{\sqrt{n}}u_{\frac{1+\gamma }{2}}\sigma , \; \overline{X} - \frac{1}{\sqrt{n}}u_{\frac{1-\gamma }{2}}\sigma \right)\right).
\]
\(\left( \overline{X} - \frac{1}{\sqrt{n}}u_{\frac{1+\gamma }{2}}\sigma , \; \overline{X} - \frac{1}{\sqrt{n}}u_{\frac{1-\gamma }{2}}\sigma \right)\)
Note the frequentist interpretation of the confidence level: it is the interval that is random, not the parameter. Upon repeated repetition of the experiment, the interval will cover the true \(\theta\) in approximately a fraction \(\gamma\) of cases:
βThe dance of confidence intervalsβ (in Geoff Cummingβs terminology): a continuous simulation for the model \(\mathscr N(\theta,1)\) with known variance, \(\theta=0\) β the true (unknown to the statistician) value. Every few moments a new exact confidence interval of level \(\gamma\) is constructed by the central statistic method, \(\overline{X}\pm\frac{1}{\sqrt n}u_{\frac{1+\gamma}{2}}\); it appears at the bottom, the whole stack shifts up, and the oldest interval leaves the screen. It is the interval that is random, not the parameter \(\theta\): from experiment to experiment it shifts and either covers \(\theta\) (in black) or does not (in red). The circular gauge on the right shows the current empirical coverage fraction β it fluctuates, but on average stays near the nominal level \(\gamma\). Move the sliders \(\gamma\) and \(n\), and use pause/resume to examine the details.
So, to find exact confidence intervals one needs to know the exact value of the probability, and hence one must use some function \(G(X, \theta )\) whose distribution does not depend on \(\theta\). Again, this function is called a central statistic, and the method of constructing an exact confidence interval based on such a random function is the central statistic method. Let us describe it in general form.
Let a random variable \(G(X, \theta )\) be such that its distribution is known and does not depend on \(\theta\). Suppose also that \(G(X, \theta )\) is strictly monotone (for example, increasing) and continuous in \(\theta\). Let \(0 \leq p_{1}<p_{2} \leq 1\) be such that \(p_{2}-p_{1}=\gamma\). For each \(i \in \{ 1,2\}\) consider the \(p_{i}\)-quantiles \(u_{p_{i}}\) of the distribution of \(G(X, \theta )\). Let \(T_{i}(X)\) be the solutions of the equations \(G\left(X, T_{i}(X)\right)=u_{p_{i}}, i \in \{ 1,2\}\). Then
\[
\mathbb {P}_{\theta }\left(T_{1}(X)<\theta <T_{2}(X)\right) = \mathbb {P}_{\theta }\left(u_{p_{1}}<G(X, \theta ) < u_{p_{2}}\right) =p_{2}-p_{1}=\gamma ,
\]
i.e. \(\left(T_{1}(X), T_{2}(X)\right)\) is a confidence interval of confidence level \(\gamma\).
In general, for a given confidence level \(\gamma \in (0,1)\), when constructing a confidence interval one can take quantiles of levels \(p_1 = \frac{1}{2}-(1-\alpha ) \gamma\), \(p_2 = \frac{1}{2} + \alpha \gamma\), where \(\alpha \in \left(1 - \frac{1}{2\gamma }, \frac{1}{2\gamma }\right)\) is arbitrary (the latter condition is imposed so that \(0 < \frac{1}{2}-(1-\alpha ) \gamma , \frac{1}{2}+\alpha \gamma < 1\)). For example, for \(\gamma = 95\%\) one could take the quantiles \((u_{1\% }, u_{96\% }), (u_{4.5\% }, u_{99.5\% })\), etc. However, to minimize the length of the confidence interval, the levels \(p_{1}\) and \(p_{2}\) in the central statistic method are often chosen symmetric about \(0.5\), since the distribution of \(G(X, \theta )\) may have heavy tails. For \(\gamma = 95\%\) these would be \(u_{2.5\% }, u_{97.5\% }\).
Let \(X_{1}, \ldots , X_{n}\) be a sample from a distribution with density \[
p_{\theta }(x)=e^{-(x-\theta )} \; \mathbb {1}_{x \geq \theta }
\] (an exponential distribution shifted by \(\theta\)). Construct a confidence interval of confidence level \(\gamma\) using the central statistic method.
Method 1. Note that \(X_{i}-\theta \sim \operatorname {Exp}(1) = \Gamma (1,1)\) (a distribution not depending on \(\theta\)). Further,
\[
\sum _{i=1}^n \left(X_{i}-\theta \right) = n\overline{X} - n\theta \sim \Gamma (n, 1), \qquad \overline{X} - \theta \sim \Gamma (n, n)
\]
Take as the central statistic \(G(X, \theta )=\sum X_{i}-n \theta\). This statistic has distribution \(\Gamma (n, 1)\), since the sum of independent, identically distributed exponential random variables has a gamma distribution. Consider the \(\frac{1-\gamma }{2}-\) and \(\frac{1+\gamma }{2}\)-quantiles \(u_{\frac{1-\gamma }{2}}\) and \(u_{\frac{1+\gamma }{2}}\) of the distribution \(\Gamma (n, 1)\). By the definition of quantiles,
\[
\gamma = \frac{1+\gamma }{2}-\frac{1-\gamma }{2}=\mathbb {P}_{\theta }\left(u_{\frac{1-\gamma }{2}}<\sum X_{i}-n \theta <u_{\frac{1+\gamma }{2}}\right).
\]
Dividing the inequality under the probability by \(n\) (so that \(\sum X_i / n = \overline{X}\)) and solving it for \(\theta\) (subtracting \(\samplemean X\) and changing sign), we get
\[
\gamma =\mathbb {P}_{\theta }\left(\overline{X}-\frac{u_{\frac{1+\gamma }{2}}}{n}<\theta <\overline{X}-\frac{u_{\frac{1-\gamma }{2}}}{n}\right) .
\]
Method 2. By means of the statistic \(X_{(1)}\). Note that for \(x \geq \theta\):
\[
\begin{align} \mathbb {P}\left(X_{1} > x\right) & = \int _x^{+\infty } e^{-(t-\theta )} dt = e^{-(x-\theta )} \\ \mathbb {P}\left(X_{(1)} > x\right) & = e^{-n(x-\theta )} \\ X_{(1)} - \theta & \sim \operatorname {Exp}(n) = \Gamma (1, n) \end{align}
\]
That is, \(X_{(1)} - \theta\) is a central statistic. Its quantile of level \(p\) equals \(u_p = -\frac{1}{n}\ln (1-p)\), whence
\[
u_{\frac{1-\gamma }{2}} = -\frac{1}{n}\ln \frac{1+\gamma }{2}, \qquad u_{\frac{1+\gamma }{2}} = -\frac{1}{n}\ln \frac{1-\gamma }{2}.
\]
Hence,
\[
\gamma = \mathbb {P}\left(u_{\frac{1-\gamma }{2}} < X_{(1)} - \theta < u_{\frac{1+\gamma }{2}}\right) = \mathbb {P}\left(-\frac{1}{n}\ln \frac{1+\gamma }{2} < X_{(1)} - \theta < -\frac{1}{n}\ln \frac{1-\gamma }{2}\right).
\]
Solving the inequality for \(\theta\) (subtracting \(X_{(1)}\) and changing sign), we get
\[
\gamma = \mathbb {P}\left(X_{(1)} + \frac{1}{n}\ln \frac{1-\gamma }{2} < \theta < X_{(1)} + \frac{1}{n}\ln \frac{1+\gamma }{2}\right)
\]
(both endpoints lie to the left of \(X_{(1)}\), which is natural: \(\theta \leq X_{(1)}\) a.s.).
Method 3. Via Chebyshevβs inequality
\[
\begin{align} \mathbb {E}\left[X_1\right] = 1 + \theta , \quad \operatorname {Var}\left[X_1\right] = 1 \\ \mathbb {P}_{\theta }\left(\left|\overline{X} - (1 + \theta )\right| \geq \varepsilon \right) < \frac{1}{\varepsilon ^2 n} \\ \mathbb {P}_{\theta }\left(\overline{X} - 1 - \varepsilon \leq \theta \leq \overline{X} - 1 + \varepsilon \right) \geq 1 - \frac{1}{\varepsilon ^2 n} = \gamma , \quad \varepsilon = \frac{1}{\sqrt{n(1-\gamma )}}\\ \end{align}
\]
Let \(X_1, \dots , X_n\) be a sample from \(U[0, \theta ], \; \theta > 0\). Construct a confidence interval for \(\theta\) of confidence level \(\gamma\), using the statistic
\(\overline{X}\);
\(X_{(1)}\);
\(X_{(n)}\).
*Note.** Each part is worth 0.5 points.
\(X_1, \ldots , X_n\) is a sample from a distribution with density \[
p_{\theta }(x) = \frac{3}{8\theta ^3}x^2 \; \mathbb {1}_{x \in [0, 2\theta ]}
\] Using the statistic \(X_{(1)}\), construct an exact confidence interval of confidence level \(\gamma\) for the parameter \(\theta\).
Let \(X_1, \ldots , X_n\) be a sample from a distribution with density \[
p_{\theta }(x) = \frac{1}{2}\exp \left(-\left|\left|x\right| - 6\theta \right|\right) \cdot \; \mathbb {1}_{\left|x\right| > 6\theta }
\] Construct a confidence interval for \(\theta > 0\) of confidence level \(\gamma = 1 - \alpha\).
\(X_1, \dots , X_n\) is a sample from the distribution of \(\xi + \eta\), where \(\xi , \eta\) are independent and have distribution \(\sim U[0, \theta ]\), \(\theta \in \Theta = (0,1)\). Construct a confidence interval for \(\theta\) of confidence level \(1 - \alpha\) using Chebyshevβs inequality.
Asymptotic Confidence Intervals
Small confidence intervals cannot always be constructed. In the case of a large sample, one can move away from the restriction on sample size and consider the asymptotic problem instead. In this setting, it is usually easier to find a small confidence interval.
So, let \(X_{1}, \ldots , X_{n}\) be a sample from \(\mathbb {P} \in \mathcal{P}=\left\{ \mathbb {P}_{\theta }, \theta \in \Theta \right\}\).
Definition 2 A sequence of intervals
\[
\left(T_1^{(n)}, \; T_2^{(n)}\right) = \left(T_1^{(n)}(X_1, \ldots , X_n), \; T_2^{(n)}(X_1, \ldots , X_n)\right)
\]
is called an asymptotic confidence interval of confidence level \(\gamma\) for \(\theta\) if for every \(\theta \in \Theta\)
\[
\varliminf _{n \to \infty } \mathbb {P}\left(\theta \in \left(T_1^{(n)}, T_2^{(n)}\right)\right) = \varliminf _{n \to \infty } \mathbb {P}\left(T_1^{(n)}(X_1, \ldots , X_n) < \theta < T_2^{(n)}(X_1, \ldots , X_n)\right) \geq \gamma ,
\]
That is, all limit points (there may in general be more than one) of the sequence \(\mathbb {P}\left(\theta \in \left(T_1^{(n)}, T_2^{(n)}\right)\right)\) lie in the interval \([\gamma , 1]\) for every \(\theta \in \Theta\).
An asymptotic confidence interval is called exact if for every \(\theta \in \Theta\) the limit
\[
\lim _{n \to \infty } \mathbb {P}\left(T_1^{(n)}(X_1, \ldots , X_n) < \theta < T_2^{(n)}(X_1, \ldots , X_n)\right) = \gamma
\]
exists.
How does one construct an asymptotic confidence interval? Let \(\widehat{\theta }_{n}\left(X_{1}, \ldots , X_{n}\right)\) be an asymptotically normal estimator of \(\theta\) with asymptotic variance \(\sigma^{2}(\theta )\), i.e.
\[
\sqrt{n} \frac{\widehat{\theta }_{n}-\theta }{\sigma (\theta )} \xrightarrow [n \to \infty ]{d_\theta } \mathscr {N}\left(0, 1\right).
\]
Consider the quantiles \(z_{\frac{1-\gamma }{2}}\) and \(z_{\frac{1+\gamma }{2}}\) of the distribution \(\mathscr {N}\left(0, 1\right)\) of levels \(\frac{1-\gamma }{2} = \frac{1}{2} - \frac{1}{2}\gamma\) and \(\frac{1+\gamma }{2} = \frac{1}{2} + \frac{1}{2}\gamma\) respectively. Then we have
\[
\lim _{n \to \infty } \mathbb {P}\left(z_{\frac{1-\gamma }{2}}<\sqrt{n} \frac{\widehat{\theta }_{n}-\theta }{\sigma (\theta )}<z_{\frac{1+\gamma }{2}}\right) = \mathbb {P}\left(z_{\frac{1-\gamma }{2}} < \xi < z_{\frac{1+\gamma }{2}}\right), \qquad \xi \sim \mathscr {N}\left(0, 1\right),
\]
and the last probability equals \(\frac{1+\gamma }{2} - \frac{1-\gamma }{2} = \gamma\). If we are fortunate with the function \(\sigma (\theta )\), then from the last equality one can extract an exact asymptotic confidence interval. If we are less fortunate, it may not be possible to obtain a confidence interval directly, but if the function \(\sigma (\theta )\) is continuous, then, by the theorem on the preservation of convergence in probability,
\[
\frac{\sigma (\theta )}{\sigma \left(\widehat{\theta }_{n}\right)} \xrightarrow [n \to \infty ]{\mathbb {P}_\theta } 1.
\]
Then, by Slutskyβs lemma,
\[
\sqrt{n} \frac{\widehat{\theta }_{n}-\theta }{\sigma \left(\widehat{\theta }_n\right)} \xrightarrow [n \to \infty ]{d_\theta } \mathscr {N}\left(0, 1\right).
\]
from which it is now easy to obtain an exact asymptotic confidence interval for \(\theta\).
Let \(X_{1}, \ldots , X_{n}\) be a sample from \(\operatorname {Bern}(\theta )\). Construct an asymptotic confidence interval of confidence level \(\gamma\) for the parameter \(\theta\).
By the central limit theorem,
\[
\sqrt{n}\frac{\overline{X} - \theta }{\sqrt{\theta (1-\theta )}} \xrightarrow [n \to \infty ]{d_\theta } \mathscr {N}\left(0, 1\right)
\]
Note that \(\overline{X}\) is a consistent estimator of \(\theta\), i.e. \(\overline{X} \xrightarrow [n \to \infty ]{\mathbb {P}_\theta } \theta , \; \forall \theta \in (0,1)\). Since the function \(\sqrt{\theta (1-\theta )}\) is continuous in \(\theta\) on \((0,1)\), we have \(\sqrt{\overline{X}(1 - \overline{X})} \xrightarrow [n\to \infty ]{\mathbb {P}_\theta } \sqrt{\theta (1 - \theta )}, \; \forall \theta \in (0,1)\). Equivalently,
\[
\frac{ \sqrt{\theta (1 - \theta )} }{\sqrt{\overline{X}(1 - \overline{X})}} \xrightarrow [n\to \infty ]{\mathbb {P}_\theta } 1 \quad \forall \theta \in (0,1).
\]
Hence,
\[
\xi \left(X,\theta \right) := \sqrt{n}\frac{\overline{X} - \theta }{\sqrt{\theta (1-\theta )}} \cdot \frac{ \sqrt{\theta (1 - \theta )} }{\sqrt{\overline{X}(1 - \overline{X})}} = \sqrt{n}\frac{\overline{X} - \theta }{\sqrt{\overline{X}\left(1-\overline{X}\right)}} \xrightarrow [n \to \infty ]{d_\theta } \mathscr {N}\left(0, 1\right)
\]
Consider the \(\frac{1-\gamma }{2}\)- and \(\frac{1+\gamma }{2}\)-quantiles \(u_{\frac{1-\gamma }{2}}\) and \(u_{\frac{1+\gamma }{2}}\) of the distribution \(\mathcal{N}(0,1)\). Then
\[
\mathbb {P}\left(u_{\frac{1-\gamma }{2}} < \xi < u_{\frac{1+\gamma }{2}}\right) \xrightarrow [n \to \infty ]{} \gamma .
\]
Further,
\[
u_{\frac{1-\gamma }{2}} < \sqrt{n}\frac{\overline{X} - \theta }{\sqrt{\overline{X}\left(1-\overline{X}\right)}} < u_{\frac{1+\gamma }{2}}
\]
is equivalent to
\[
\overline{X}-\sqrt{\frac{\overline{X}\left(1-\overline{X}\right)}{n}} u_{\frac{1+\gamma }{2}} < \theta < \overline{X}-\sqrt{\frac{\overline{X}\left(1-\overline{X}\right)}{n}} u_{\frac{1-\gamma }{2}}
\]
(note that the larger quantile ends up on the left side of the inequality, and the smaller one on the right). Set
\[
T_1^{(n)} := \overline{X}-\sqrt{\frac{\overline{X}\left(1-\overline{X}\right)}{n}} u_{\frac{1+\gamma }{2}}, \qquad T_2^{(n)} := \overline{X}-\sqrt{\frac{\overline{X}\left(1-\overline{X}\right)}{n}} u_{\frac{1-\gamma }{2}}
\]
Then we have \(\mathbb {P}\left( T_{1}^{(n)}<\theta <T_{2}^{(n)}\right) \xrightarrow [n \to \infty ]{} \gamma\). Hence, \(\left(T_{1}^{(n)}, T_{2}^{(n)}\right)\) is an asymptotic confidence interval of confidence level \(\gamma\).
\(\left(T^{(n)}_1, T^{(n)}_2\right)\), where \[
T_1^{(n)} := \overline{X}-\sqrt{\frac{\overline{X}\left(1-\overline{X}\right)}{n}} u_{\frac{1+\gamma }{2}}, \qquad T_2^{(n)} := \overline{X}-\sqrt{\frac{\overline{X}\left(1-\overline{X}\right)}{n}} u_{\frac{1-\gamma }{2}}
\]
Let \(X_1, \dots , X_n\) be a sample from the shifted Cauchy distribution, i.e. \[
\begin{align} p_{\theta }(x) = \frac{1}{\pi (1+ (x - \theta )^2 )} \end{align}
\] Construct an asymptotic confidence interval for \(\theta\) of confidence level \(\alpha\).
Let \(X_1, \dots , X_n\) be a sample from \(\mathrm{Pois}(\theta )\). Construct an asymptotic confidence interval for \(\theta\) of confidence level \(\alpha\).
Let \(X_1, \dots , X_n\) be a sample from \(\Gamma (\theta , \lambda ), \; \theta , \lambda > 0\). Construct an asymptotic confidence interval for \(\lambda\) of confidence level \(\alpha\), if
\(\theta\) is known (1 point);
\(\theta\) is unknown (2 points).
Recall that if \(X_1 \sim \Gamma (\theta , \lambda )\), then \[
\begin{align} p_{X_1}(x) & = \frac{\lambda ^\theta }{\Gamma (\theta )} x^{\theta - 1}e^{-\lambda x} \; \mathbb {1}_{ x \geq 0}, \\ \mathbb {E}\left[X_1\right] & = \frac{\theta }{\lambda }, \quad \operatorname {Var}\left[X_1\right] = \frac{\theta }{\lambda ^2}, \quad \mathbb {E}\left[X_1^k\right] = \frac{1}{\lambda ^k}\theta (\theta + 1)(\theta +2)\cdot \dots \cdot (\theta + k - 1), \; k \in \mathbb {N} \end{align}
\]