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For an absolutely continuous $f$ with $f(0)=0$, this causal Abel integral equation has inversionUnder Laplace-transform growth hypotheses, the convolution theorem for Laplace transforms converts the original kernel into $\Gamma (1-\alpha )s^{\alpha -1}$. Since $\mathcal{L}[f^{\prime }]=s\mathcal{L}[f]$, the transformed inverse is $s^{-\alpha }\mathcal{L}[f^{\prime }]/\Gamma (1-\alpha )$, giving the formula. It also holds locally in the corresponding integrable convolution class: convolving the two power kernels produces the constant $B(\alpha ,1-\alpha )=\Gamma (\alpha )\Gamma (1-\alpha )$ by the beta-gamma identity, so substitution recovers ${\int }_0^t{f}^{\prime }=f(t)$. The same convolution identity proves uniqueness in that class. For example $f(t)=t$ gives $\Phi =t^{\alpha }/[\Gamma (\alpha +1)\Gamma (1-\alpha )]$.

The beta distribution is an absolutely continuous probability distribution for a random variable $X$ between zero and one. It can describe uncertainty about a probability, such as an unknown coin's chance of heads. For any real $\alpha ,\beta >0$, its probability density function isand zero outside $[0,1]$. The normalizer is the Euler beta function, related to the gamma function by the beta-gamma identity:Thus the probability density function integrates to one. ^[ref].

The endpoint behavior explains why the restrictions are exactly $\alpha ,\beta >0$: near zero the probability density function behaves like a constant times $x^{\alpha -1}$, and near one like a constant times $(1-x{)}^{\beta -1}$. These improper integrals converge for positive parameters, even when the probability density function diverges at zero for $\alpha <1$ or at one for $\beta <1$. There is no probability mass at either endpoint: $\mathbb{P}(X=0)=\mathbb{P}(X=1)=0$. A probability density function's height is not a probability; an interval's probability is the area under it.

Some useful examples of the beta distribution are:

The arcsine distribution curve is plotted only for $0.001\le x\le 0.999$; its finite plotted ends do not represent finite endpoint values of the probability density function. Using the inverse sine, the shaded central interval has exact probability $\mathbb{P}(0.1<X<0.9)=1-4\arcsin (\sqrt{0.1})/\pi \approx 0.5903$, calculated analytically rather than by treating the cropped curve as normalized.

To separate location from spread, write $\kappa =\alpha +\beta$ and $m=\alpha /\kappa$. The mean and variance of the beta distribution areThe mean specifies the balance between the parameters; their sum $\kappa$ controls concentration about that mean. Multiplying both parameters by the same positive number preserves the mean and changes the variance. Larger $\kappa$ puts more area near $m$ and makes the probability density function taller there; the total area stays one. Small $\kappa$ allows much greater spread, sometimes favoring the endpoints rather than the mean itself.

For example, keeping mean $m=0.3$ and taking $\kappa =5,20,80$ gives the beta distributions $\mathrm{Beta}(1.5,3.5)$, $\mathrm{Beta}(6,14)$, and $\mathrm{Beta}(24,56)$. Their variances are $0.035$, $0.01$, and $0.21/81\approx 0.00259259$, respectively. Parameters need not be integers.

The formulas follow from two short moment calculations. Multiplying the probability density function by $x$ or $x^2$ increases the first parameter of the Euler beta function:Apply the Gamma function recurrence to the beta-gamma identity, then subtract $(\mathbb{E}[X]{)}^2$ from $\mathbb{E}[X^2]$ to obtain the variance.

The beta distribution is especially convenient as a prior distribution for the unknown success probability $P$ of Bernoulli trials. The Beta-binomial conjugacy says that $h$ successes and $t$ failures update a prior distribution $\mathrm{Beta}(\alpha ,\beta )$ to a posterior distribution $\mathrm{Beta}(\alpha +h,\beta +t)$. For example, suppose a coin has prior distribution $\mathrm{Beta}(2,2)$ and we observe eight heads and two tails. Its posterior distribution is then $\mathrm{Beta}(10,4)$, with posterior mean $5/7$. ^[ref].

The Beta function obeysThese identities follow immediately from the beta-gamma identity and the Gamma function recurrence.

The beta-gamma identity gives $\Gamma (z/2{)}^2=B(z/2,z/2)\Gamma (z)$ for $\mathrm{Re}z>0$. A zero of $\Gamma (z)$ would therefore force zeros at every $z/2^m$. But $w\Gamma (w)=\Gamma (1+w)\to 1$ as $w\to 0$ in the right half-plane, a contradiction. The Gamma function recurrence then extends nonvanishing to every point where the meromorphic gamma function is finite. Nonpositive integers are poles, not zeros.

For real $p>0$ and $0<s<p$, the displayed improper integral converges: its endpoint powers are $x^{s-1}$ near zero and $x^{s-p-1}$ near infinity. The change of variables $t=x^p$ givesHere the beta function, the beta-gamma identity, and the Gamma reflection formula evaluate the integral. The same formula holds for complex $s$ in the Mellin fundamental strip $0<\mathrm{Re}s<p$, by absolute convergence and analytic continuation within the strip. For integer $p$, a sector contour of angle $2\pi /p$ gives a direct residue theorem proof, with a consistent branch of $z^{s-1}$.

Set $t=x^2$, so $dx=dt/(2\sqrt{t})$. ThenThe beta-gamma identity, the Gamma function recurrence and the gamma reflection formula give

For $\mathrm{Re}z>0$, absolute convergence permits the two-dimensional gamma integralPut $u=rs$, $v=r(1-s)$, with $r>0$, $0<s<1$ and Jacobian determinant $r$. Separation of the integrals gives the beta-gamma identityThe beta integral is finite in this half-plane. Thus a zero of $\Gamma (z)$ would imply $\Gamma (z/2)=0$, and iteration would imply $\Gamma (z/2^m)=0$ for every positive integer $m$. But integration by parts gives $w\Gamma (w)=\Gamma (1+w)$, and continuity near one makes $\Gamma (1+w)\to 1$ as $w\to 0$. It cannot vanish for sufficiently small $w$, a contradiction. Therefore the Gamma function has no zeros in the right half-plane.

For any $z$ outside the nonpositive integers, choose an integer $n$ with $\mathrm{Re}(z+n)>0$. The Gamma function recurrence givesAt the excluded integers gamma has poles, with nonzero residues by the same recurrence. Hence the meromorphic gamma function has no zeros anywhere; at its poles it has no finite value to which a nonzero-value assertion could apply.

In the integral defining the Beta function, put $t=(1+x)/2$ and then $u=x^2$. Symmetry givesThis proves the identity for $\mathrm{Re}z>0$. Both sides extend by analytic continuation to the same meromorphic function, equivalently by the beta-gamma identity and the Gamma duplication formula. The identity holds meromorphically for every $z\in \mathbb{C}$; both sides are finite for $z\notin \{0,-1,-2,\dots \}$ and have matching poles at the excluded points.

With $u=x^4$, the change of variables formula givesThe beta-gamma identity, $\Gamma (1/2)=\sqrt{\pi }$, and the Gamma reflection formula giveTherefore

Multiply the two defining Gamma integrals. For $p,q>0$ the nonnegative product integrand allows Tonelli's theorem:Use $r=x+y$, $u=x/(x+y)$, so $x=ru$, $y=r(1-u)$ and the positive quadrant becomes $r>0$, $0<u<1$. The Jacobian determinant has absolute valueAfter including this factor, the transformed integral separates:This proves the beta-gamma identity,All changes of variables are made inside the open positive domains; their boundaries do not contribute to the integrals.

Treat the left side as a causal convolution with $t^{-\alpha }$. Under the usual Laplace-transform hypotheses, the convolution theorem for Laplace transforms givesFor an absolutely continuous $f$ with $f(0)=0$, the derivative transform is $\mathcal{L}[f^{\prime }]=s\hat{f}$. ConsequentlySince $\mathcal{L}[t^{\alpha -1}]=\Gamma (\alpha )s^{-\alpha }$, inverting the product provesHere $0<\alpha <1$ makes both power kernels locally integrable. For example assume exponential growth bounds sufficient for the transforms and a regularity class where the convolution is defined. This is the Abel integral equation of fractional order; the initial value removes the extra boundary term in the derivative transform.

For $f(t)=t$, the remaining integral is ${\int }_0^t(t-\tau {)}^{\alpha -1}d\tau =t^{\alpha }/\alpha$. ThusSubstitution into the original equation checks the result: scaling $\tau =tu$ gives $tB(\alpha +1,1-\alpha )/[\Gamma (\alpha +1)\Gamma (1-\alpha )]=t$ by the beta-gamma identity.

Conditional on $\theta$, the future Bernoulli random variable is independent of the observed statistical sample and has success probability $\theta$. Its posterior predictive probability is therefore the posterior mean of $\theta$. Put $A=a+s$ and $B=b+n-s$. Thenusing the beta-gamma identity and the Gamma function recurrence. HenceThe whole posterior predictive distribution of $X_{n+1}$ is the Bernoulli distribution with this success probability.