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The cumulative distribution function isThe total integral of the nonnegative probability density function is one.

Normalization of the probability density function requiresHence $\boxed{c=t-1}$. The distribution is a Pareto distribution with minimum one and shape parameter $t-1$.

An expected value of a signed random variable is finite precisely when its absolute value is integrable. With the standard normal distribution probability density function,For $\alpha <1/2$, this integral equals $1/[\sqrt{2\pi }(1/2-\alpha )]<\infty$. For $\alpha \ge 1/2$, its integrand does not decay sufficiently and the integral diverges. ThereforeThe value is zero by the odd function and the symmetric distribution. For the other $\alpha$, both positive and negative parts have infinite integrals; formal cancellation would give only a Cauchy principal value, not an expected value. This distinction is the integrability threshold for exponential powers of a Gaussian.

On the product support $[0,1{]}^2$, the joint probability density function factors asThe equality also holds outside the support with both sides zero. The probability density function factorization criterion proves that $U$ and $V$ are independent random variables.

A joint probability density function integrates to one. HeresoThis value also makes the probability density function nonnegative everywhere.

By independence, the likelihood function is the product of the individual probability density functions:Writing $A(\mu )={\sum }_i|x_i-\mu |$, the log-likelihood function is $\ell =n\ln \lambda -n\ln 2-\lambda A(\mu )$. Here $\lambda$ is a rate, the reciprocal of the scale in the usual Laplace distribution notation.

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 cumulative distribution function must tend to one at infinity, so $\boxed{k=1}$. Differentiation gives the probability density function $f_X(x)=e^{-x}$ for $x>0$ and zero for $x<0$; there is no atom at zero. This is an exponential distribution of rate one. Its first two moments are ${\int }_0^{\infty }x{e}^{-x}dx=1$ and ${\int }_0^{\infty }{x}^2{e}^{-x}dx=2$, so

The probability density function is normalized because ${\int }_0^{\infty }\alpha e^{-\alpha t}dt=1$. Integration by parts gives $\mathbb{E}[T]={\int }_0^{\infty }t\alpha e^{-\alpha t}dt=1/\alpha$ and $\mathbb{E}[T^2]=2/{\alpha }^2$. ThereforeThese are the mean and variance of the exponential distribution with rate $\alpha$.

The function is a probability density function, rather than a cumulative distribution: its integral over both intervals must be one. Thus $2\alpha +2\alpha =1$ and

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:

For an integrable discrete random variable, its expectation and variance are $\mu ={\sum }_{n\in \mathbb{Z}}n{p}_n$ and ${\sum }_{n\in \mathbb{Z}}(n-\mu {)}^2{p}_n$, respectively. The first formula requires $\sum |n|p_n<\infty$ for a finite ordinary mean; the variance may be infinite. A probability density function gives $P(\alpha <Y<\beta )={\int }_{\alpha }^{\beta }f(y)dy$.

Conditioning on $X$ and using independence givesThe interchange is justified by Tonelli's theorem, since all integrands are nonnegative. Consequently $Z$ has an absolutely continuous distribution, withIts integral is ${\sum }_n{p}_n=1$. This is the convolution of independent random variables, specifically a Gaussian smoothing of an integer-valued distribution for the given normal density. The specified weights are normalized because $2(3/{\pi }^2){\sum }_{n\ge 1}{n}^{-2}=1$.

Conditioned on being inside the circle, the point is a uniform random point in a disk of radius $1/2$. The annulus from $r$ to $r+dr$ has area $2\pi rdr+O(d{r}^2)$. Dividing by the disk's area givesThe radial probability density function is $8r$ on $[0,1/2]$ and integrates to one. Without conditioning, the corresponding annulus probability for a point in the original square is $2\pi rdr+O(d{r}^2)$.

For a continuous random variable with probability density function $p$, its expected value and variance areThe mean requires an integrable first absolute moment, and a finite variance requires a finite second moment. When these hold, $\mathrm{Var}X=\mathbb{E}[X^2]-(\mathbb{E}X{)}^2$.

Write $\bar{s}>0$ for the mean in minutes. An exponential distribution with that mean has rate $1/\bar{s}$, so its probability density function isThe density has units of inverse minutes, ensuring that its integral is a dimensionless probability.

Interpreting $f$ as the probability density function of the random variable, for $\alpha \le \beta$ one hasA single point has zero probability, so including or excluding either endpoint does not change this result.

Normalize the probability density function on its nonnegative support:The integral follows from one integration by parts, with the boundary term zero since $\lambda >0$. Hence $A={\lambda }^2$. This is a gamma distribution of shape $2$ and rate $\lambda$.

The probability density function is nonnegative for $\alpha >0$. Its integral over the support isso it is normalized for every $\alpha >0$. This is the unit-threshold Pareto distribution.

Integrating the probability density function from its lower endpoint to $x$ gives the cumulative distribution functionIt is continuous and strictly increasing on the support. Hence the unique median $m$ satisfies $F(m)=1/2$, or $m^{-\alpha }=1/2$. Therefore