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The prior mean $1/(\beta +1)$ decreases as $\beta$ increases. More strongly, its cumulative distribution function and upper-tail probability areFor every fixed $t$, that upper-tail probability decreases with $\beta$. The statistician increasingly favors smaller values of $\theta$; as $\beta \to \infty$, the prior distribution concentrates near zero. For $\beta >1$, its variance $\beta /[(\beta +1{)}^2(\beta +2)]$ also decreases. This is concentration of a beta prior near zero.

With the data fixed, the derivatives areThus the prior weight increases and the posterior mean decreases. Their limits are $\lambda \to 1$ and ${\tilde{\theta }}_1\to 0$. Both the changing prior mean and the increasing weight contribute to the change in the posterior mean.

Put $M_0=\alpha +x_0\beta$. As a linear combination of independent normal random variables under the posterior distribution,The shortest 95% credible interval isHere $\Phi$ is the standard normal distribution cumulative distribution function. To justify shortest length, consider all intervals of a fixed length $d$. If $f$ is this normal distribution posterior density, the interval probability ${\int }_a^{a+d}f(t)dt$ has derivative $f(a+d)-f(a)$ with respect to $a$. The normal distribution is symmetric and its density strictly decreases with distance from its mean, so this probability is uniquely maximized when the midpoint is $m_0$. The centered interval can therefore attain 95% posterior probability with the least length. It is both an equal-tailed credible interval and a highest posterior density interval.

The target is the experiment's mean at $x_0$. A posterior predictive distribution for a new noisy observation would instead have variance $v_0+{\sigma }^2$; that extra observation variance does not belong in the requested interval.

An unknown variance requires a joint prior distribution for $(\alpha ,\beta ,{\sigma }^2)$. The analysis is not determined until a prior for the unknown variance, and its relation to the coefficients, is specified. A maximum-likelihood estimate substituted for ${\sigma }^2$ does not by itself account for its posterior uncertainty.

For example, retain the independent coefficient prior distributions from part (b) and independently assign $s={\sigma }^2$ an inverse-gamma distribution with shape $a_0>0$ and scale $b_0>0$, whose probability density function is proportional to $s^{-a_0-1}{e}^{-b_0/s}$. Then the joint posterior density is proportional toConditional on $s$, the two coefficient posterior distributions are still the independent normal distributions in part (b), with $s$ substituted for ${\sigma }^2$. The other full conditional distribution isThe marginal coefficient posterior distribution requires marginalization over $s$; it is generally nonnormal and the two coefficients need not remain independent. In particular, the posterior cumulative distribution function of the requested mean becomesIts shortest 95% credible interval must be found from this marginal posterior distribution, for example by numerical integration or posterior simulation. This is normal regression with independent coefficient and variance priors.

A different, convenient choice is a normal-inverse-gamma prior, in which the conditional coefficient prior covariance matrix is proportional to ${\sigma }^2$. That choice gives a multivariate Student's t-distribution after marginalization, so the interval uses its univariate Student's t-distribution quantiles. It changes the coefficient prior specification; a Student interval does not follow automatically from retaining the fixed-${\tau }_0,{\tau }_1$ priors in part (b).

The derivative of the log ratio is $\frac{1}{2}(n/q-1)$, so $\lambda (q)$ increases from zero to one on $(0,n)$ and decreases to zero on $(n,\infty )$. Rejecting for a sufficiently small ratio therefore gives two tails with $0<q_1<n<q_2$.

Under $H_0$, Cochran's theorem gives the exact sampling distribution $Q=S/{\sigma }_0^2\sim {\chi }_{n-1}^2$, independently of the unknown mean. Writing $F_{n-1}$ for its cumulative distribution function, the two cutoffs are determined byThe first equation calibrates the exact size of a statistical test. The second is $\lambda (q_1)=\lambda (q_2)$, so both tails belong to one likelihood ratio cutoff. Hence the exact critical region is $\boxed{\{q\le q_1\}\cup \{q\ge q_2\}}$. Its two tail probabilities are not generally equal: equal-tailed chi-squared cutoffs give a different test.

The full parameter space has dimension two and the null space dimension one. The regular Wilks theorem therefore gives $W\stackrel{d}{\to }{\chi }_1^2$ under $H_0$ as $n\to \infty$. If $F_1$ denotes the cumulative distribution function of the chi-squared distribution with one degree of freedom, the approximate size-$\alpha$ rejection region isThe corresponding approximate p-value isThe one degree of freedom here describes the large-sample likelihood-ratio test statistic; it is different from the exact residual degrees of freedom $n-1$ in the next part.

An equal-tailed credible interval for a continuous posterior distribution has probability $\alpha /2$ below its lower endpoint and $\alpha /2$ above its upper endpoint. If $F$ is the posterior cumulative distribution function, the endpoints are $F^{-1}(\alpha /2)$ and $F^{-1}(1-\alpha /2)$.

A highest posterior density interval is a credible interval consisting of points of highest posterior density: every included point has density at least as high as every excluded point, apart from null-set and boundary choices. It is a posterior highest density region when that region is an interval. For a continuous unimodal density with two interior endpoints, their densities agree. A mode at a support boundary can instead force one endpoint to that boundary; a multimodal posterior can have a disconnected highest-density region rather than one interval. Equal tails and highest density need not give the same interval.

The exponential distribution prior has rate $\lambda =1/3$ and support $\mu \ge 0$, so its prior density is $(1/3)e^{-\mu /3}{1}_{\{\mu \ge 0\}}$. By Bayes' theorem,where $C$ is a normalizing constant independent of $\mu$.

Completing the square gives a useful exact reference for the numerical sampling. PutThen this normal mean posterior with an exponential prior is a normal distribution with location $m$ and variance $s^2$, truncated below at zero. With $\varphi$ the standard normal distribution density and $\Phi$ its cumulative distribution function,The exponential rate is $1/3$, not three; the prior mean is three. The support restriction must be retained in both the formula and the sampler.

Let $W=U^2$. On $0\le U\le 1$, squaring preserves order. Therefore for $0\le w\le 1$,Including the tails gives the cumulative distribution functionDifferentiating on the interior of the support gives the probability density functionThe divergence near zero is integrable: ${\int }_0^1(2\sqrt{w}{)}^{-1}dw=1$. There is no point mass at zero, and the assigned value of a probability density function at a single endpoint does not affect the probability distribution.

Assume $a<b$. An increasing affine map sends the two endpoints to zero and one when $ca+d=0$ and $cb+d=1$. Solving givesIndeed, for $0\le t\le 1$, the cumulative distribution function of the transformed random variable iswhich is the cumulative distribution function of the continuous uniform distribution on $[0,1]$.

A decreasing affine map works too, by swapping the endpoints:These are the two possible choices, since the image interval must have endpoints zero and one. The case $a=b$ cannot be transformed into a nondegenerate continuous uniform distribution by deterministic constants.

Complete the square in the transformation:Since $0<X<2$ almost surely, $0<Y\le 1$. For $0\le y<1$, put $r=\sqrt{1-y}$. The event $Y>y$ is equivalent to $1-r<X<1+r$. Integrating the given probability density function over both branches at once gives the survival functionThus, including the tails,The cumulative distribution function on $(0,1)$ is $1-\sqrt{1-y}$. Differentiation givesBoth inverse branches matter: $x=1\pm \sqrt{1-y}$. The endpoint divergence is integrable, and $\mathbb{P}(Y=1)=\mathbb{P}(X=1)=0$, so no endpoint atom is needed.

Integrating the probability density function from the lower endpoint gives the cumulative distribution functionThe cumulative distribution function is continuous at $x=1$ and tends to one at infinity.

Differentiate the cumulative distribution function for $y>0$ using the chain rule and the standard normal density:Thus a probability density function isThe value assigned at zero is immaterial. The probability density function is unbounded near zero, but $F_Y(0)=0$ and continuity of the cumulative distribution function there show that there is no point mass. This is a gamma distribution with shape $1/2$ and rate $1$.

The random variable $Y=Z^2/2$ is nonnegative, so its cumulative distribution function is zero for $y<0$. If $y\ge 0$, thenThe last step uses the symmetry of the standard normal distribution, $\Phi (-z)=1-\Phi (z)$. Hence

Since $X\ge 0$, the function $X\mapsto X^2$ is strictly increasing on its support. For $y\ge 0$,and $F_Y(y)=0$ for $y<0$. Differentiating the cumulative distribution function givesThe density value at the single point $y=0$ is immaterial. Hence $Y$ has an exponential distribution with rate $1/2$. Direct integration by parts, or its standard expected value, yields

The cumulative distribution function is continuous and strictly increasing for $x>0$, so the median is the unique positive $c$ satisfying $F_X(c)=1/2$. Consequently $e^{-c^2/2}=1/2$, givingThere is no atom at $c$, so $\mathbb{P}(X>c)=1-F_X(c)=1/2$ as required.

Since $X$ is nonnegative, $F_X(x)=0$ for $x<0$. For $x\ge 0$, use $\frac{d}{dt}{e}^{-t^2/2}=-t{e}^{-t^2/2}$ to obtainThus the cumulative distribution function isIn particular it tends to one as $x\to \infty$, confirming normalization of this unit-scale Rayleigh distribution.

The cumulative distribution function isFor an absolutely continuous probability distribution, the probability density function is a nonnegative integrable function $f_X$ withThus interval probabilities are integrals of $f_X$, and $F_X^{\prime }=f_X$ almost everywhere, in particular at continuity points of $f_X$.

The word “continuous” in the source is understood in this density sense. A merely continuous cumulative distribution function need not possess a probability density function; it can instead describe a singular continuous distribution.

Reversing sign reverses the inequality. Since $X$ has a probability density function, it has no point masses, soDifferentiating the cumulative distribution function givesalmost everywhere: the minus sign outside $F_X$ cancels the minus sign from the chain rule. If point masses were allowed, the cumulative distribution function would instead require the left limit $1-F_X((-y{)}^{-})$.

Translate the inequality inside the cumulative distribution function:Differentiating with the chain rule gives the probability density functionalmost everywhere. Translation shifts the probability distribution without changing its shape or its total probability.

For any real $y$, put $k=\lfloor y\rfloor$. Since $Y$ is integer-valued,Continuity of the cumulative distribution function of $X$ therefore givesThe strict inequality $X<k+1$ matters before using the absence of point masses. The resulting probability distribution is discrete, with probability mass functionThus $Y$ has no ordinary probability density function with respect to length on the real line. Its cumulative distribution function is a step function; its derivative is zero almost everywhere and does not recover its probabilities. The floor function has converted a density into masses on the integers.