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Computing the same expected value directly from the Poisson distribution instead givesFor each fixed positive integer $n$, the ratio test shows that this sum converges. This identity is Dobiński's formula. The case $n=0$ also gives $B(0)=1$ if $0^0$ is interpreted as one in this particular moment formula.

Expand the square and use linearity of expectation, together with $\mathbb{E}[X]=\mu$ and $\mathbb{E}[1]=1$:This proof applies whenever the second moment is finite, and does not require the random variable to be continuous.

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.

The second moment isUsing the mean $\mathbb{E}[X]=1$, the variance is therefore

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

The time-independent Schrödinger equation isPut $\alpha =m\omega /\hslash$. For $\psi =e^{-\alpha x^2/2}$, differentiation gives ${\psi }^{\prime \prime }=({\alpha }^2{x}^2-\alpha )\psi$. Substitution cancels the terms in $x^2$ and leaves $E_0={\hslash }^2\alpha /(2m)=\hslash \omega /2$.

The Gaussian integral gives $\int e^{-\alpha x^2}dx=\sqrt{\pi /\alpha }$. Hence a normalized wave function isMultiplication by any constant of modulus one gives the same quantum state. The position probability density is $|{\psi }_0{|}^2=\sqrt{\alpha /\pi }{e}^{-\alpha x^2}$. Its first moment vanishes by oddness. Integration by parts gives $\int x^2{e}^{-\alpha x^2}dx=(2\alpha {)}^{-1}\int e^{-\alpha x^2}dx$, with the boundary term zero. Therefore

The moment-generating function of a real random variable $X$ isfor real $t$ for which this expected value is finite. Because the integrand is nonnegative, the extended expected value is always defined, but it may equal $+\infty$. We always have $M_X(0)=1$; the usual uniqueness and convergence theorems require finiteness on an open interval containing zero. When this condition holds, derivatives at zero recover the moments, $M_X^{(r)}(0)=\mathbb{E}[X^r]$.

The strong law of large numbers says that, for real independent and identically distributed random variables $Z_1,Z_2,\dots$ with $\mathbb{E}|Z_1|<\infty$,The assertion is almost sure convergence of the entire sequence of sample means. Finite first absolute moment suffices; a finite variance is not required.