Continuous uniform distribution ($U(a,b)$)

A continuous uniform distribution has constant density on its interval of support.

The cumulative distribution function of $U^a$ on $[0,1]$ is $z^{1/a}$. Differentiating gives density $z^{1/a-1}/a$ on $(0,1)$, which is the beta distribution probability density function with parameters $(1/a,1)$. This also gives a simple instance of inverse transform sampling: to sample a density $\alpha z^{\alpha -1}$ on $(0,1)$, raise a uniform draw to the power $1/\alpha$.

For independent random variables $X,Y$ with the uniform distribution on $(0,b)$, use the probability density of a product of random variables and integrate $1/(b^{2}x)$ over $z/b<x<b$. The displayed probability density function is zero off $(0,b^2)$ and has integral one. More generally, for independent and identically distributed factors $U_j$ with this uniform distribution, $W_n={\prod }_{j=1}^n{U}_j$ satisfiesThe expectation of a product of independent random variables proves these identities directly.

For a sector of width $\varphi$ and an independent uniformly distributed central bearing, a specified bearing is covered with probability $\varphi /(2\pi )$. Two antipodal bearings cannot both be covered when $\varphi <\pi$. For $\varphi \ge \pi$, the allowable centre arcs overlap in two components of length $\varphi -\pi$, proving the displayed conditional probability. Averaging over the width's probability density function gives unconditional coverage probabilities.

For independent $U,V$ uniform on $[0,1]$, real roots occur on $U\le V^2$, a region of probability $1/3$. On that region both roots are nonpositive. Both lie in $[-1,0]$ exactly when $\max (0,2V-1)\le U\le V^2$. Its area is $1/12$, so the conditional probability of both absolute values being at most one is $1/4$. Joint probability density integration turns algebraic discriminant and root constraints into simple planar areas.

A uniform random variable has a uniform distribution. Independent such variables can supply independent inverse transform sampling; reusing a single variable for two transforms generally creates dependence.

A Steinhaus random variable is uniformly distributed on the complex unit circle. It can be generated as $ϵ\cos \theta +i\delta \sin \theta$ using independent signs $ϵ,\delta$ and an independent uniform angle $\theta \in [0,\pi /2]$. This conditional sign representation transfers Rademacher sum inequalities to random complex phases.

For independent Steinhaus random variables, condition on their acute angles and use the sharp Rademacher second-moment inequality on the $2d$ sign coefficients $a_i\cos {\theta }_i,i{a}_i\sin {\theta }_i$. Their squared moduli sum to ${\sum }_i|a_i{|}^2$ independently of the angles, so the conditional first moment is at least the same fixed square-root bound. Averaging proves the assertion.

The square of a uniform distribution variable on $(0,1)$ has cumulative distribution function $\sqrt{x}$ and an integrable inverse-square-root probability density function. Reflection replaces $x$ by $1-x$ in that density.

For independent $U_1,U_2\sim \mathrm{Uniform}(0,1)$, the sum has triangular densityIts upper tail for $1\le c\le 2$ is $(2-c{)}^2/2$.

Let $X\ge 0$ have density $g$, let $U$ be independent and uniform on $[0,1]$, and setThen

Write $h(y)={\int }_y^{\infty }g(t)/tdt$. The two pieces in a uniform split are independent exactly whenSetting $z=0$ shows that $h(y)=c{e}^{-cy}$ for some $c>0$. Thus the pieces are independent exactly when they are independent exponential variables of the same rate; equivalently, the total has gamma density $c^{2}x{e}^{-cx}$.