Affine Recodings of Bernoulli Outcomes

Why Recoding Changes the Values, Not the Uncertainty

A Bernoulli variable records one of two outcomes, commonly coded as 0 for failure and 1 for success. In applications, the quantity of interest may instead be a score, cost, payoff, or measurement that assigns different numerical values to those outcomes. An affine recoding expresses this change as a shift and rescaling: $Y=a+bX$. Its possible values and probabilities follow directly from those of $X$, while its moments show exactly how the recoding affects location and variability.

Throughout, let $X\sim\operatorname{Bernoulli}(p)$, so $P(X=1)=p$ and $P(X=0)=1-p$, with $0\leq p\leq 1$. Let $a,b$ be fixed real numbers, and define $Y=a+bX$. The constant $a$ shifts both outcomes; $b$ determines their separation and, if negative, reverses their order.

Definition: An affine recoding of $X$ is a random variable $Y=a+bX$ for fixed real constants $a$ and $b$. If $b\ne 0$ and $0<p<1$, its two possible values are $a$ and $a+b$, with probabilities $1-p$ and $p$, respectively. The support consists of values having positive probability. If $p$ is 0 or 1, or if $b=0$, some or all of these values coincide or have probability zero, so the support must be reduced accordingly.

Distribution and Two-Point Support

The outcome $X=0$ maps to $Y=a$, and the outcome $X=1$ maps to $Y=a+b$. Therefore, when $b\ne 0$ and $0<p<1$, the probability mass function is

f_Y(y)=(1-p)\mathbf{1}\{y=a\}+p\mathbf{1}\{y=a+b\},

where $\mathbf{1}\{A\}$ equals 1 when the condition $A$ holds and 0 otherwise. This expression specifies the mass at each of the two points and is zero elsewhere. The distribution function can be written without splitting into cases according to the sign of $b$:

F_Y(y)=P(Y\leq y) =(1-p)\mathbf{1}\{a\leq y\}+p\mathbf{1}\{a+b\leq y\}.

If $b>0$, then $a<a+b$, so the lower value has probability $1-p$ and the upper value has probability $p$. If $b<0$, then $a+b<a$: the lower value now has probability $p$, and the upper value has probability $1-p$. The probabilities remain attached to the original outcomes $X=1$ and $X=0$, not to whichever value happens to be larger.

Theorem: For $Y=a+bX$, with $X\sim\operatorname{Bernoulli}(p)$, every integer raw moment of order $r\geq 0$ is $$ E[Y^r]=(1-p)a^r+p(a+b)^r. $$ For every integer $r\geq 1$, the central moment is $$ E[(Y-E[Y])^r] =b^r\left[(1-p)(-p)^r+p(1-p)^r\right]. $$ In particular, $E[Y]=a+bp$ and $\operatorname{Var}(Y)=b^2p(1-p)$.

These formulas apply for all real $a,b$ and $0\leq p\leq 1$, including degenerate cases. For $r=0$, the raw-moment formula gives $E[1]=1$, including when a possible value is zero, since $y^0$ here denotes the constant function 1.

Deriving the Moment Formulas

Because $X$ takes only the values 0 and 1, $Y$ takes the corresponding values $a$ and $a+b$. Weighting each value by its probability gives the raw moment:

$$ E[Y^r] =(1-p)a^r+p(a+b)^r. $$

In particular,

$$ E[Y]=(1-p)a+p(a+b)=a+bp. $$

To obtain the central moments, subtract this mean from each possible value. When $X=0$, $Y-E[Y]=a-(a+bp)=-bp$. When $X=1$, $Y-E[Y]=a+b-(a+bp)=b(1-p)$. Thus

E[(Y-E[Y])^r] =(1-p)(-bp)^r+p\bigl(b(1-p)\bigr)^r =b^r\left[(1-p)(-p)^r+p(1-p)^r\right].

Taking $r=2$ yields

\operatorname{Var}(Y) =b^2\left[(1-p)p^2+p(1-p)^2\right] =b^2p(1-p)\left[p+(1-p)\right] =b^2p(1-p).

This derivation shows why a shift does not affect variance, while multiplying by $b$ multiplies variance by $b^2$. The sign of $b$ changes the direction of the recoding but not its variance. The derivation also gives higher central moments when needed; for example, the third central moment is $b^3p(1-p)(1-2p)$. $\square$

Worked Example: A Net Payoff From a Bernoulli Outcome

A clinical decision has probability $p=0.30$ of success. The decision incurs a fixed cost of 2 units whether or not it succeeds, and success earns 10 units. Let $X=1$ denote success and $X=0$ failure. Find the distribution, mean, and variance of the net payoff $Y$.

The payoff is $-2$ on failure and $10-2=8$ on success. Thus $Y=-2+10X$, so $a=-2$ and $b=10$. The support points and masses are

P(Y=-2)=0.70,\qquad P(Y=8)=0.30.

The mean is

$$ E[Y]=a+bp=-2+10(0.30)=1. $$

Direct calculation checks the result: $0.70(-2)+0.30(8)=-1.4+2.4=1$. The variance is

\operatorname{Var}(Y)=b^2p(1-p)=10^2(0.30)(0.70)=21.

A second check uses $E[Y^2]=0.70(4)+0.30(64)=2.8+19.2=22$, so $\operatorname{Var}(Y)=E[Y^2]-(E[Y])^2=22-1=21$.

Worked Example: A Negative Rescaling Reverses the Support Order

Let $X\sim\operatorname{Bernoulli}(0.80)$, where 1 denotes success, and define $Z=5-2X$. Find the support, probability masses, mean, and variance.

If $X=0$, then $Z=5$, with probability $0.20$. If $X=1$, then $Z=3$, with probability $0.80$. Because the slope is negative, success maps to the smaller value:

P(Z=3)=0.80,\qquad P(Z=5)=0.20.

The mean formula gives $E[Z]=5+(-2)(0.80)=3.4$. Direct calculation agrees: $0.80(3)+0.20(5)=2.4+1=3.4$. The variance is

\operatorname{Var}(Z)=(-2)^2(0.80)(0.20)=0.64.

Although the outcome labels have reversed order, the mass associated with each original Bernoulli outcome has not changed. In particular, the probability at 3 is $p$, not $1-p$.

Worked Example: Recoding a Bernoulli Variable as Two Scores

Suppose $X\sim\operatorname{Bernoulli}(0.40)$, and a scoring rule assigns 20 points for $X=0$ and 50 points for $X=1$. Find the distribution and first two raw moments of the score $S$.

The difference between the scores is $50-20=30$, so $S=20+30X$. Its support is $\{20,50\}$, and

P(S=20)=0.60,\qquad P(S=50)=0.40.

The mean is $E[S]=20+30(0.40)=32$. The second raw moment is

$$ E[S^2] =(0.60)(20^2)+(0.40)(50^2) =(0.60)(400)+(0.40)(2500) =240+1000=1240. $$

Consequently, $\operatorname{Var}(S)=1240-32^2=1240-1024=216$. This also follows from $30^2(0.40)(0.60)=900(0.24)=216$.

Degenerate Cases and a Useful Check

The two-point description requires two distinct values with positive probability. If $b=0$, then $Y=a$ with probability 1, whatever $p$ is. If $p=0$, then $X=0$ with probability 1 and $Y=a$; if $p=1$, then $Y=a+b$ with probability 1. In each case the variance formula correctly gives zero. A formula listing $a$ and $a+b$ should not be mistaken for two support points when one of the values is impossible or when the values coincide.

Worked Example: Checking a Boundary Probability

Let $X\sim\operatorname{Bernoulli}(1)$ and $Y=7-3X$. Since $X=1$ with probability 1, $Y=7-3=4$ with probability 1. Thus the support is the singleton $\{4\}$, $E[Y]=4$, and $\operatorname{Var}(Y)=0$. The general formulas confirm this: $7-3(1)=4$ and $(-3)^2(1)(0)=0$.

Common Mistake: Assigning Probabilities by Numerical Order

A frequent error is to assume that the larger recoded value must have probability $p$, or that a negative multiplier changes $p$ to $1-p$. Neither is correct. The probability $p$ remains attached to the event $X=1$, whose recoded value is $a+b$; the probability $1-p$ remains attached to $X=0$, whose recoded value is $a$. Only after identifying those mappings should the two values be arranged from smaller to larger, if that is useful.

Another check is to distinguish a change in location from a change in spread. Adding a constant changes the mean but not the variance. Multiplying by $b$ changes the variance by a factor of $b^2$, not $b$. This remains true for negative $b$, because variance is based on squared deviations.

Key takeaway: For $Y=a+bX$ with $X\sim\operatorname{Bernoulli}(p)$, the values $a$ and $a+b$ inherit probabilities $1-p$ and $p$. The mean is $a+bp$, and the variance is $b^2p(1-p)$. Check the sign of $b$ and any boundary probabilities before describing the support.

Check Your Understanding

Answer each question by tracking the original outcomes $X=0$ and $X=1$.

If $X\sim\operatorname{Bernoulli}(0.25)$ and $Y=4+6X$, what are the support points and their probabilities?
For the variable in Question 1, calculate $E[Y]$ and $\operatorname{Var}(Y)$.
If $X\sim\operatorname{Bernoulli}(0.60)$ and $Z=8-5X$, which support point has probability $0.60$? Explain.
What is the support of $Y=a+bX$ when $p=0$, and what is its variance?
Why does a negative value of $b$ not make $\operatorname{Var}(Y)$ negative?