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For a probability density function $P$, the mean and variance areThese formulas assume the required moments are finite; finiteness of the second moment ensures a finite mean as well.

The triangular probability density function is normalized because $2{\int }_0^1(1-x)dx=1$. It is even, so the mean integrand is odd and $\mu =0$. Its variance is consequently its second moment:

The variance is the second moment minus the square of the expected value:

Using the sum of squares, the second moment is

The density is symmetric about $7/2$, giving expected value $\mathbb{E}[X]=7/2$. Its second moment is $\mathbb{E}[X^2]=\frac{1}{4}({\int }_1^3{x}^{2}dx+{\int }_4^6{x}^{2}dx)=89/6$. ThereforeEquivalently, each equally weighted interval has variance $1/3$ and its centre is $3/2$ from the overall mean, giving $1/3+9/4$ by the law of total variance.

Write $\lambda =\ln ({\rho }^{-1})>0$. The probability density function normalizes when $1=B{\int }_0^{\infty }{e}^{-\lambda x}dx=B/\lambda$, so $B=\lambda$. Its survival function is $e^{-\lambda x}$ for $x\ge 0$; hence, for $x>y\ge 0$,Differentiating this conditional tail givesThe value assigned at the single endpoint does not affect any probability. The memorylessness of the exponential distribution says that $X-y$, conditional on $X>y$, is again an exponential distribution of rate $\lambda$. Its first two moments are $1/\lambda$ and $2/{\lambda }^2$, found by integration by parts, so translation leaves the variance unchanged:

When the relevant moments are finite, the mean and variance areEquivalently, ${\sigma }^2=\int x^{2}f(x)dx-{\mu }^2$. The mean requires an integrable absolute first moment; finite second moment guarantees both quantities are finite.

Repeated integration by parts gives, for every nonnegative integer $j$,With $A={\lambda }^2$, the first two moments are $\mathbb{E}[X]={\lambda }^2(2!/{\lambda }^3)=2/\lambda$ and $\mathbb{E}[X^2]={\lambda }^2(3!/{\lambda }^4)=6/{\lambda }^2$. Subtracting the square of the mean givesIf $X$ carries units, $\lambda$ has inverse units of $X$, so the mean and standard deviation have the units of $X$ and the variance has their square.

The second moment is $E[X^2]=\alpha {\int }_1^{\infty }{x}^{1-\alpha }dx$, finite exactly for $\alpha >2$, where it equals $\alpha /(\alpha -2)$. Subtracting the square of the mean givesFor $1<\alpha \le 2$, the mean exists but the variance is infinite. For $0<\alpha \le 1$, the ordinary centered variance is not defined because the mean is not finite. Thus there is no finite variance for any $\alpha \le 2$.

The three preceding probabilities satisfyTheir expected value and second moment areSubtracting the squared mean givesThe value $\mu =2(2/10)$ also follows from linearity of expectation. The variance includes the finite-population reduction caused by sampling without replacement.

The second moment isSubtracting the squared expected value gives the variance and standard deviation:The variance is nonnegative because $0<r\le R$. Both $\mu$ and $\sigma$ have units of time. At $r=R$ the result reduces to $R/\sqrt{12}$, the standard deviation of a uniform distribution over one full cycle.

The expected value is not defined: the integral of $yf(y)$ over the positive half-line diverges logarithmically to $+\infty$, and the negative half-line gives $-\infty$. In particular, symmetry alone does not supply a mean. More explicitly,The second moment also diverges because $y^{2}f(y)\to x_0/\pi >0$ as $|y|\to \infty$. ThusThe symmetric Cauchy principal value of the first-moment integral is zero, but it is not an expected value and does not define a variance.

For the uniform distribution, the expected value and second moment areHence the variance is

For attractive nearest-neighbour interactions, $J>0$, the transformed model is a ferromagnetic Ising model. In three dimensions and at $T<T_c$, its infinite-volume spontaneous magnetization has two branches. As the effective field crosses zero, the equilibrium mean spin changes discontinuously from a negative value to a positive value. Since $(\partial h/\partial \mu {)}_T=1/2$, varying the physical chemical potential crosses this same first-order phase transition.

The particle-hole-symmetric coexistence line isThe mean Ising spin is a rescaled occupation density. Writing $n=⟨N⟩/V$ for the physical number density,The negative-spin phase is a dilute gas and the positive-spin phase is a dense, liquid-like phase. If the spontaneous spin magnitude at coexistence is $m_0(T)>0$, their number densities areThus the physical order parameter is a density or occupation imbalance, not an actual magnetic moment of these particles. In a closed system with an intermediate mean density, the same transition appears as phase coexistence between the two densities. At $T=T_c$ the density contrast vanishes continuously; above $T_c$, sweeping the chemical potential gives no coexistence jump. A true discontinuity requires the thermodynamic limit; a finite sum over finitely many cells remains analytic.

Let each step have length $a$, and let $K$ be the number of right steps among $N$ independent steps. Then $K$ has a binomial distribution and the physical displacement is $x=a[K-(N-K)]=a(2K-N)$. Therefore $K=(N+x/a)/2$, and the position probability mass isThis is zero unless $x/a$ is an integer, $|x/a|\le N$, and $N+x/a$ is even. The parity condition matters because the walk alternates between the two sublattices. If distance is measured in lattice spacings, replace $x/a$ by $x$.

Normalization follows from the binomial theorem. Its first two moments provide useful checks:These are exact discrete-time statements; a smooth continuum probability density requires coarse-graining over the lattice and its parity oscillation.

For $N$ with Poisson distribution of mean $\lambda$, the equality $n\mathbb{P}(N=n)=\lambda \mathbb{P}(N=n-1)$ proves the identity by summation. For an independent random sum of independent claims $S$ with this count, exchangeability of the summands gives $\mathbb{E}[Sh(S)]=\lambda \mathbb{E}[Xh(S+X)]$, where $X$ on the right is independent of $S$ and has the severity law. These identities hold for nonnegative measurable functions, and for integrable signed functions. They obtain moments without assuming that a positive exponential moment exists.

For positive claims with a finite positive moment-generating function neighbourhood, use $e^u>1+u+u^2/2$ and $e^u>1+u+u^2/2+u^3/6$ for $u>0$ in the defining adjustment coefficient equation. They imply the displayed strict upper bounds, where $r_2$ is the positive quadratic root. Its defining polynomial is negative at zero and positive at $r_1$, proving $r_2<r_1$. Higher exponential-series truncations similarly yield upper bounds from more raw moments.

A posterior moment is a moment computed under a posterior distribution. Its first raw moment is the posterior mean; its second central moment is the posterior variance.

A power option is a European contingent claim whose payoff is a fixed power of the terminal underlying price. In the Black-Scholes model, its value is $S_t^p\exp ((p-1)r(T-t)+\frac{1}{2}p(p-1){\sigma }^2(T-t))$ when the required moments are finite.

On a probability space, the same inequality readsIf the displayed moments are finite, the product is an integrable random variable. The random variables can be dependent: the estimate controls their possible overlap without assuming a factorization of the expected value. This is useful when a calculation needs the product to be integrable but the joint probability distribution is complicated.