Beta Reflection and Complement Transformations

Why Complementation Matters

A Beta random variable takes values between zero and one, so its complement $1-X$ is also a natural quantity to study. If $X$ represents a proportion, then $1-X$ represents the remaining proportion. The transformation is especially useful when a probability is awkward to calculate in one tail but straightforward in the other, or when a moment of the complement is needed.

The key fact is that complementation swaps the two shape parameters: if $X$ has a Beta distribution with parameters $a$ and $b$, then $1-X$ has a Beta distribution with parameters $b$ and $a$. We will derive this fact from the density, then use it to connect symmetry, probabilities, and moments.

Definition: For $a>0$ and $b>0$, the distribution $\operatorname{Beta}(a,b)$ has density $$ f_X(x)=\frac{x^{a-1}(1-x)^{b-1}}{B(a,b)},\qquad 0<x<1, $$ and density zero outside $(0,1)$. Here $$ B(a,b)=\int_0^1 x^{a-1}(1-x)^{b-1}\,dx $$ is the beta function. The parameters $a$ and $b$ are positive shape parameters.

The Distribution of the Complement

Set $Y=1-X$. This is a one-to-one transformation from $(0,1)$ to $(0,1)$, with inverse $x=1-y$ and absolute derivative $|dx/dy|=1$. Applying the one-dimensional change-of-variables formula gives the density of $Y$.

Theorem: If $X\sim\operatorname{Beta}(a,b)$, where $a,b>0$, then $$ 1-X\sim\operatorname{Beta}(b,a). $$

To see every part of the transformation, substitute $x=1-y$ into the density of $X$:

\begin{aligned} f_Y(y) &=f_X(1-y)\left|\frac{d}{dy}(1-y)\right|\\ &=\frac{(1-y)^{a-1}y^{b-1}}{B(a,b)},\qquad 0<y<1. \end{aligned}

The beta function is unchanged when its arguments are interchanged. Indeed, substituting $u=1-x$ into its defining integral gives $B(a,b)=B(b,a)$. Therefore,

f_Y(y)=\frac{y^{b-1}(1-y)^{a-1}}{B(b,a)},\qquad 0<y<1,

which is exactly the $\operatorname{Beta}(b,a)$ density. The Jacobian is one, but it is still important to include it in the transformation: for a different transformation, omitting its absolute derivative could give an incorrect density. $\square$

Worked Example: Identify the Complement Distribution

Suppose $X\sim\operatorname{Beta}(2,5)$, and let $Y=1-X$. Identify the distribution of $Y$.

The complement theorem swaps the two shape parameters. The first parameter of $X$ is $2$, and the second is $5$, so $$ Y=1-X\sim\operatorname{Beta}(5,2). $$

For a direct density check, the density of $X$ is $30x(1-x)^4$ on $(0,1)$, since $B(2,5)=1/30$. Thus $$ f_Y(y)=f_X(1-y)=30(1-y)y^4=30y^4(1-y),\qquad 0<y<1, $$ which is the $\operatorname{Beta}(5,2)$ density. The larger shape parameter has moved from the factor near zero to the factor near one in the original variable; after complementation, those roles reverse.

Symmetry and Equal Shape Parameters

A distribution on $(0,1)$ is symmetric about $1/2$ when reflecting values around $1/2$ leaves its distribution unchanged. For a Beta variable, reflection around $1/2$ is exactly the complement transformation. The theorem therefore gives both a simple symmetry condition and its converse.

Theorem: If $X\sim\operatorname{Beta}(a,b)$ with $a,b>0$, then $X$ is symmetric about $1/2$ if and only if $a=b$.

If $a=b$, the density itself verifies symmetry: $$ f_X(1-x) =\frac{(1-x)^{a-1}x^{a-1}}{B(a,a)} =f_X(x),\qquad 0<x<1. $$ Equivalently, the complement theorem gives $1-X\sim\operatorname{Beta}(a,a)$, the same distribution as $X$.

For the converse, suppose the distribution is symmetric about $1/2$, so $X$ and $1-X$ have the same distribution. The Beta mean is $E[X]=a/(a+b)$, and the complement theorem gives $E[1-X]=b/(a+b)$. Equality in distribution implies equal means, so $$ \frac{a}{a+b}=\frac{b}{a+b}. $$ Since $a+b>0$, this implies $a=b$. Thus no unequal pair of positive shape parameters can produce a symmetric Beta distribution.

When $a=b$, the mean is $1/2$, but a mean of $1/2$ alone does not establish symmetry. The symmetry follows from the full density identity, not merely from the mean.

Worked Example: Use Symmetry to Compute a Tail

Let $X\sim\operatorname{Beta}(3,3)$. Find $P(X>0.7)$ in terms of a lower-tail probability.

The shape parameters are equal, so $X$ and $1-X$ have the same distribution. Then $$ P(X>0.7)=P(1-X<0.3)=P(X<0.3). $$ The inequalities reverse because subtracting from one reverses their order. Since the Beta distribution is continuous, using $<$ or $\le$ at $0.3$ gives the same probability. Thus an upper-tail probability above $0.7$ equals the lower-tail probability below $0.3$; no numerical evaluation of either probability is required to establish the equality.

Transforming CDFs and Probabilities

Write $F_X(x)=P(X\le x)$ for the CDF of $X$, and $F_Y(y)=P(Y\le y)$ for $Y=1-X$. A Beta distribution has a density and hence no point masses. For $0\le y\le1$,

\begin{aligned} F_Y(y) &=P(1-X\le y)\\ &=P(X\ge 1-y)\\ &=1-F_X(1-y). \end{aligned}

The last equality uses continuity: $P(X\ge c)=1-P(X<c)=1-F_X(c)$ when $P(X=c)=0$. Since $Y\sim\operatorname{Beta}(b,a)$, the same identity can be written $$ F_{b,a}(y)=1-F_{a,b}(1-y),\qquad 0\le y\le1, $$ where $F_{a,b}$ denotes the CDF of a $\operatorname{Beta}(a,b)$ variable.

The same order reversal converts events involving intervals and tails. For $0\le c\le1$, $$ P(X>c)=P(Y<1-c)=F_Y(1-c), $$ where continuity again makes the strict or non-strict inequality immaterial. More generally, if $0\le r<s\le1$, then $$ r<X<s\quad\Longleftrightarrow\quad 1-s<Y<1-r. $$ Both endpoints change, and their order reverses.

Worked Example: Convert an Upper Tail to a Lower Tail

Let $X\sim\operatorname{Beta}(2,5)$. Find $P(X>0.8)$ by using the complement.

Set $Y=1-X$, so $Y\sim\operatorname{Beta}(5,2)$. The event transforms as $$ X>0.8\quad\Longleftrightarrow\quad Y<0.2. $$ The $\operatorname{Beta}(5,2)$ density is $30y^4(1-y)$. Its CDF on $[0,1]$ is obtained by integrating: $$ F_Y(y)=\int_0^y 30t^4(1-t)\,dt =6y^5-5y^6. $$ Consequently, $$ \begin{aligned} P(X>0.8) &=P(Y<0.2)\\ &=6(0.2)^5-5(0.2)^6\\ &=\frac{6}{3125}-\frac{5}{15625}\\ &=\frac{30-5}{15625} =\frac{1}{625}. \end{aligned} $$ The result is $0.0016$. The small value is consistent with the $\operatorname{Beta}(2,5)$ density concentrating toward zero.

Worked Example: Transform an Interval Probability

Suppose $X\sim\operatorname{Beta}(3,2)$. Express $P(0.25<X<0.7)$ as a probability for the complement, and evaluate it.

For $Y=1-X$, the distribution is $\operatorname{Beta}(2,3)$. The interval endpoints reverse: $$ 0.25<X<0.7\quad\Longleftrightarrow\quad 0.3<Y<0.75. $$ The density of $Y$ is $12y(1-y)^2$. Integrating it gives $$ F_Y(y)=\int_0^y12t(1-t)^2\,dt=6y^2-8y^3+3y^4. $$ Therefore, $$ \begin{aligned} P(0.25<X<0.7) &=F_Y(0.75)-F_Y(0.3)\\ &=\left(6(0.75)^2-8(0.75)^3+3(0.75)^4\right)\\ &\quad-\left(6(0.3)^2-8(0.3)^3+3(0.3)^4\right)\\ &=0.94921875-0.3483\\ &=0.60091875. \end{aligned} $$ This calculation illustrates why both interval endpoints must be complemented: the interval $(0.25,0.7)$ becomes $(0.3,0.75)$, not $(0.25,0.3)$ or $(0.7,0.75)$.

Moments of the Complement

The complement theorem identifies the entire distribution of $1-X$, so moments can be obtained from the Beta moments with the parameters exchanged. In particular, for a nonnegative integer $k$, the raw $k$th moment of $Y=1-X$ is $$ E[Y^k]=E[(1-X)^k]=\frac{(b)_k}{(a+b)_k}, $$ where $(c)_k=c(c+1)\cdots(c+k-1)$ for $k\ge1$, and $(c)_0=1$. This formula also follows directly by integrating the density: $$ E[(1-X)^k] =\frac{1}{B(a,b)}\int_0^1 x^{a-1}(1-x)^{b+k-1}\,dx =\frac{B(a,b+k)}{B(a,b)}. $$ Repeated use of the beta-function recurrence $B(u,v+1)=\frac{v}{u+v}B(u,v)$ gives the stated product. Here $a,b>0$ and $k$ is a nonnegative integer.

For the first two moments, the consequences are especially useful: $$ E[1-X]=\frac{b}{a+b},\qquad \operatorname{Var}(1-X)=\frac{ab}{(a+b)^2(a+b+1)}. $$ The mean changes from $a/(a+b)$ to $b/(a+b)$, while the variance stays the same. More generally, because $1-X$ is an affine function of $X$, $$ \operatorname{Var}(1-X)=\operatorname{Var}(X),\qquad \operatorname{Cov}(X,1-X)=-\operatorname{Var}(X). $$ The variance identity follows from $\operatorname{Var}(c+X)=\operatorname{Var}(X)$ and $\operatorname{Var}(-X)=\operatorname{Var}(X)$; the covariance identity follows from $\operatorname{Cov}(X,1-X)=\operatorname{Cov}(X,1)-\operatorname{Cov}(X,X)$.

Worked Example: Compare the First Two Moments

Let $X\sim\operatorname{Beta}(2,3)$, and let $Y=1-X$. Find the means and variances of both variables, then find $E[Y^2]$.

The complement theorem gives $Y\sim\operatorname{Beta}(3,2)$. The means are $$ E[X]=\frac{2}{2+3}=\frac25,\qquad E[Y]=\frac{3}{3+2}=\frac35. $$ For either variable, the product of the shape parameters is $6$, and the sum is $5$. Hence $$ \operatorname{Var}(X)=\operatorname{Var}(Y) =\frac{6}{5^2(5+1)} =\frac{6}{150} =\frac1{25}. $$ Finally, for $Y\sim\operatorname{Beta}(3,2)$, $$ E[Y^2]=\frac{3(3+1)}{(3+2)(3+2+1)} =\frac{12}{30} =\frac25. $$ Checking against the mean and variance gives the same value: $$ E[Y^2]=\operatorname{Var}(Y)+(E[Y])^2 =\frac1{25}+\frac9{25} =\frac{10}{25} =\frac25. $$

Common Pitfalls and Why the Transformation Helps

The complement transformation is a compact way to reuse Beta distribution results, but it is easy to apply it incompletely. Keep the event transformation separate from the parameter transformation: the parameters swap because the density changes, while the event endpoints reverse because the map $x\mapsto1-x$ is decreasing.

Leaving the parameters in the original order. If $X\sim\operatorname{Beta}(a,b)$, its complement is $\operatorname{Beta}(b,a)$, not $\operatorname{Beta}(a,b)$ in general.
Changing only one interval endpoint. The interval $r<X<s$ becomes $1-s<1-X<1-r$; both endpoints are complemented and their order reverses.
Assuming every Beta distribution is symmetric. Symmetry about $1/2$ requires $a=b$. The fact that a variable lies between zero and one does not imply that its density is symmetric.
Equating symmetry with a centered mean. A mean of $1/2$ is not, by itself, a proof of symmetry. For Beta distributions, symmetry follows from equal shape parameters and the reflected density.

In applications, a probability about a proportion near one can often be recast as a probability about its complement near zero. A moment of the remaining proportion can likewise be obtained by swapping the Beta parameters rather than expanding $(1-X)^k$ and combining several moments. These are two uses of the same distributional identity.

Check Your Understanding

Try these questions using the complement theorem and event transformations.

If $X\sim\operatorname{Beta}(4,7)$, what is the distribution of $1-X$?
For which positive values of $a$ and $b$ is $\operatorname{Beta}(a,b)$ symmetric about $1/2$? Give a reason, not just a condition on its mean.
If $X\sim\operatorname{Beta}(2,5)$, rewrite $P(0.4<X<0.9)$ as an interval probability for $Y=1-X$, and state the distribution of $Y$.
For $X\sim\operatorname{Beta}(a,b)$, what are $E[1-X]$ and $\operatorname{Var}(1-X)$?
Explain why $P(X>c)=P(1-X<1-c)$, and why the strict inequality does not affect the probability for a Beta random variable.