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The standard normal density isIt is nonnegative, and the Gaussian integral gives ${\int }_{-\infty }^{\infty }\varphi (z)dz=1$, so it is a probability density function. Completing the square gives the moment-generating function of a standard normal variableThus $\mathbb{E}Z=M_Z^{\prime }(0)=0$ and

Normalization of the probability density function givessoThe expected value isThe second moment isTherefore the variance is

For the median $m$,so

A Bayesian inverse problem consists of a prior distribution ${\mu }_0$ on the unknown $u\in X$, a reference measure on the data space $Y$, and a jointly measurable likelihood $L(u;y)$ such that $L(u;\cdot )$ is a probability density function for ${\mu }_0$-almost every $u$. For observed data $y$, Bayes' theorem defines the posterior distribution byprovided $0<Z(y)<\infty$.

The following four assumptions are sufficient, with statements understood for ${\mu }_0$-almost every $u$ and every $y$:

The first two assumptions give $0<Z(y)<\infty$. The last two allow the dominated convergence theorem to prove $L^1({\mu }_0)$ continuity of the normalized posterior density, which is equivalent to continuity in total variation distance.

Expand the exponent in the Inverse Gaussian distribution with unit shape parameter:Consequently its probability density function can be written aswhereIndeed, $-A(\eta )=\sqrt{-2\eta }=1/\mu$. This is the canonical form of a one-parameter exponential family, and its natural parameter is therefore

Here the natural statistic is $T(X)=X$. The exponential-family derivative identities giveandThus

At fixed intrinsic ratio $q$, part a givesSince $\mu$ is uniform, the change-of-variables formula for a probability density yields the conditional probability density functionThe inverse-square-root singularity at $Q=q$ is integrable, and direct integration gives one.

For an intrinsic probability density function $\mathcal{P}(q)$, the law of total probability averages the conditional density from part c. An object observed with ratio $Q$ can only have $q\le Q$, soThis is an Abel transform of $\mathcal{P}(q)/\sqrt{1-q^2}$. Its normalization follows by reversing the order of integration and using ${\int }_q^1\mathcal{P}(Q\mid q)dQ=1$.

Withthe factor $\sqrt{1-q^2}$ cancels from the Abel transform:HenceThis probability density function is normalized and again favors rounder projections.

Let $(Y_n)$ be independent and identically distributed random variables with the standard Cauchy distribution, whose probability density function isIts first absolute moment diverges sinceOn the other hand, its characteristic function is $\phi (t)=e^{-|t|}$. The characteristic function of a sum of independent variables therefore givesThus the stability of the Cauchy distribution implies that $S_n/n$ has the standard Cauchy distribution for every $n$. In particular,where $Z$ is standard Cauchy and $\stackrel{d}{\to }$ denotes convergence in distribution.

Let $N$ be the homogeneous Poisson point process in ${\mathbb{R}}^3$ with intensity measure $\lambda dx$, and map each point $x$ to its radius $g(x)=|x|$. For an interval $(a,b]\subseteq {\mathbb{R}}_{+}$, the pushforward measure of the intensity isThe Mapping theorem for Poisson point processes therefore shows that the radii form a Non-homogeneous Poisson point process with intensity function

If $R$ is the distance to the closest star, the event $\{R>r\}$ says that the ball of radius $r$ contains no points. The zero-count probability of a Poisson distribution givesDifferentiating its cumulative distribution function yields the probability density function

The two-dimensional Fokker-Planck equation iswith Fokker-Planck probability currentThe point initial condition iswhere $\delta$ is the Dirac delta function. The reflecting boundary condition for a diffusion iswith $\mathbf{n}$ the outward unit normal. This zero-flux condition conserves the integral of the probability density function over the square.

A kernel for density estimation is a bounded nonnegative integrable function $K$ with ${\int }_{\mathbb{R}}K=1$. Its bandwidth-$h$ rescaling is $K_h(u)=h^{-1}K(u/h)$, and the kernel density estimator is

Fix $x$ and put $V_i=K_h(x-X_i)$. Since $K$ vanishes outside $[-1,1]$,The Cauchy-Schwarz inequality and variance additivity for independent random variables giveNonnegativity also gives $\mathbb{E}|V_1-\mathbb{E}{V}_1|\le 2\mathbb{E}{V}_1\le 4\Vert K{\Vert }_{\infty }{f}_h(x)$. Taking the better estimate at each $x$ and applying Tonelli theorem proves

For nonnegative $a,b$ and $0\le \theta \le 1$, $\min a,b\le a^{\theta }{b}^{1-\theta }$. Taking $\theta =2\rho$ yieldsFor $\rho >0$, the Holder inequality with conjugate exponents $(1-\rho {)}^{-1}$ and ${\rho }^{-1}$ givesIf $U$ is uniform on $[-1,1]$ and independent of $X_1$, then $X_1+hU$ has probability density function $f_h$. Sincethe last integral is at most $(1+h{)}^{\delta }\int (1+|x|{)}^{\delta }f(x)dx$. Substitution proves the second displayed bound. The case $\rho =0$ is the first bound integrated using $\int f_h=1$ and follows directly.

For a random variable with a continuous probability distribution and probability density function $f_X$, the mean and variance areA finite mean requires $\int |x|f_X(x)dx<\infty$. A finite variance additionally requires the second moment to be finite; without that condition the nonnegative integral for the variance may be infinite.

Normalization of the probability density function and the specified mean give two equations:Dividing the second by the first gives $2b/3=1$. ConsequentlyBoth are positive, as required for the stated probability density function.

The probability density function must integrate to one, soAlso, linearity of expectation and the mean and variance of the standard normal distribution giveUsing the probability density function to compute this expected value givesThese are the values $\Gamma (1/2)$ and $\Gamma (3/2)$ of the gamma function.

Differentiate the cumulative distribution function for $y>0$, using ${\Phi }^{\prime }(z)=(2\pi {)}^{-1/2}{e}^{-z^2/2}$ and $d\sqrt{2y}/dy=1/\sqrt{2y}$. The probability density function isIts value at zero can be assigned arbitrarily without changing any probability. Although this probability density function diverges as $y\downarrow 0$, $F_Y(0)=0$ and there is no point mass at zero. Equivalently, $Y$ has a gamma distribution with shape $1/2$ and rate one.

Be careful about the index: the probability density function already contains a factor $x$, so $\mathbb{E}[X_i^r]=I_{r+1}$. The recurrence relation at $k=1$ yields $I_3=\theta I_1=\theta$. Therefore linearity of expectation givesThus $\hat{\theta }$ is an unbiased estimator.

For the rate-one exponential distribution, the probability density function is $e^{-t}$ for $t\ge 0$. By integration by parts,Thus

Differentiating the cumulative distribution function in part (ii) gives the probability density function $n{t}^{n-1}$ on $(0,1)$. HenceThus an unbiased estimator based on the sample maximum is

For a random variable with a probability density function, a singleton has zero probability:Equivalently, its cumulative distribution function is continuous, so it has no jump at $x$.