Shifting and Scaling Discrete Uniform Supports

Why Transforming the Support Matters

A discrete uniform variable assigns the same probability to each integer in a finite consecutive set. A transformed variable, such as a shifted measurement or a code obtained by multiplying the original value, can have a different-looking support. The probabilities are straightforward only if the support is identified correctly: multiplying by an integer greater than one creates gaps, and multiplying by a negative integer reverses the order of the possible values.

The essential method is to map every possible value of the original variable through the transformation. Because a nonzero multiplier maps distinct values to distinct values, no transformed value combines the probability of multiple original values. That fact is what preserves equal probabilities.

The Original Model and Its Transformation

Definition: Let $X$ have a discrete uniform distribution on the consecutive integers $\{m,m+1,\ldots,n\}$, where $m$ and $n$ are integers with $m\le n$. There are $N=n-m+1$ possible values, and $$ P(X=x)=\frac{1}{N}, \qquad x=m,m+1,\ldots,n. $$ For any other value $x$, $P(X=x)=0$.

Consider the affine transformation $Y=a+bX$, where $a$ is a fixed constant and $b$ is a nonzero integer. The value $a$ shifts the transformed values; $b$ scales them. If $a$ is an integer, all transformed values are integers. If $a$ is not an integer, the same support and probability statements apply, but the transformed values need not be integers.

Theorem: If $X$ is uniform on $\{m,\ldots,n\}$ and $Y=a+bX$ with $b\ne 0$, then $$ \operatorname{supp}(Y)=\{a+bx:x=m,m+1,\ldots,n\}, \qquad P(Y=y)=\frac{1}{n-m+1} $$ for each $y$ in that support. For $y$ outside the support, $P(Y=y)=0$. When $b>0$, the smallest and largest support values are $a+bm$ and $a+bn$. When $b<0$, they are $a+bn$ and $a+bm$, respectively.

The set notation is important: if $|b|>1$, the support does not contain every integer between its smallest and largest values. It advances in steps of $|b|$, in increasing or decreasing order according to the sign of $b$. A shift alone, $Y=X+c$, is the special case $b=1$: it preserves the spacing and the probabilities, while moving every support value by $c$.

Why the Probabilities Stay Equal

The original variable has $N$ possible values, each with probability $1/N$. For $b\ne 0$, the transformation is one-to-one: if $a+bx_1=a+bx_2$, then $b(x_1-x_2)=0$, so $x_1=x_2$. Thus each possible value of $Y$ comes from exactly one possible value of $X$. In particular, no transformed value collects probability from two original values.

P(Y=y)=P\left(X=\frac{y-a}{b}\right)=\frac{1}{N}

This expression applies when $(y-a)/b$ is one of the integers $m,m+1,\ldots,n$. Otherwise, $y$ is not in the transformed support and its probability is zero. This condition prevents the common error of assigning positive probability to values between the transformed support points.

List the original support.
Write every integer from $m$ through $n$, and count them using $N=n-m+1$.

Transform each support value.
Calculate $a+bx$ for each possible $x$. This direct mapping is reliable even when $b$ is negative.

Assign probabilities and check events.
Give each distinct transformed value probability $1/N$. For an event, identify which original support values map into it, then count those values.

Worked Examples

Worked Example: A Positive Scale Creates Gaps

A hypothetical coding system selects $X$ uniformly from the integers $4,5,\ldots,10$, then reports $Y=2X+3$. Find the support and probability mass function of $Y$, and calculate $P(15\le Y\le21)$.

Method. Map the original support through the formula, then use the one-to-one property to assign equal probabilities. There are $$ N=10-4+1=7 $$ original values, so each has probability $1/7$.

Transform the values one at a time: $$ \begin{array}{c|rrrrrrr} x&4&5&6&7&8&9&10\\ \hline 2x+3&11&13&15&17&19&21&23 \end{array} $$ Therefore, $\operatorname{supp}(Y)=\{11,13,15,17,19,21,23\}$, and $P(Y=y)=1/7$ for each listed value. Every other value has probability zero.

For $15\le Y\le21$, the included support values are $15,17,19,21$, four values. Hence $$ P(15\le Y\le21)=4\left(\frac17\right)=\frac47\approx0.5714. $$ As a check, the corresponding original values are $X=6,7,8,9$, also four of the seven equally likely values.

Common mistake avoided: Treating every integer from 11 through 23 as possible would incorrectly include values such as 12 and 22. The scale factor of 2 means the support advances by 2, not by 1.

Worked Example: A Negative Scale Reverses the Support

A hypothetical rating $X$ is uniform on $\{1,2,3,4,5,6\}$. A converted score is $Y=20-3X$. Find the support, $P(Y=11)$, and $P(Y\le8)$.

Method. Apply the transformation to each original value. This avoids incorrectly using the lower original endpoint to find the lower transformed endpoint when the multiplier is negative.

There are $6-1+1=6$ equally likely original values, each with probability $1/6$. The mapping is $$ \begin{array}{c|rrrrrr} x&1&2&3&4&5&6\\ \hline 20-3x&17&14&11&8&5&2 \end{array} $$ Thus the support, written in increasing order, is $\{2,5,8,11,14,17\}$. Each of those six values has probability $1/6$, and other values have probability zero.

The value $Y=11$ occurs when $X=3$, so $P(Y=11)=1/6$. The event $Y\le8$ includes $Y=8,5,2$, three support values. Therefore, $$ P(Y\le8)=\frac{3}{6}=\frac12=0.5000. $$ The original values producing this event are $X=4,5,6$, again three of the six possible values, which checks the calculation.

Common mistake avoided: Since the multiplier is negative, $Y$ decreases as $X$ increases. The smallest value is $20-3(6)=2$, not $20-3(1)=17$.

Worked Example: An Interval Event Under a Negative Scale

A hypothetical measurement $X$ is uniform on the integers from $-3$ through $5$. A reported index is $W=7-2X$. Find the support and $P(6\le W\le14)$.

Method. First count and map the original possibilities. There are $$ N=5-(-3)+1=9 $$ values, each with probability $1/9$. Applying $W=7-2X$ gives $$ \begin{array}{c|rrrrrrrrr} x&-3&-2&-1&0&1&2&3&4&5\\ \hline 7-2x&13&11&9&7&5&3&1&-1&-3 \end{array} $$

So the support in increasing order is $\{-3,-1,1,3,5,7,9,11,13\}$. Each listed value has probability $1/9$.

To find the event probability, solve its inequalities while respecting the negative coefficient: $$ 6\le7-2X\le14. $$ Subtracting 7 gives $-1\le-2X\le7$. Dividing by $-2$ reverses both inequality directions, giving $$ -\frac72\le X\le\frac12. $$ The original support values meeting these bounds are $-3,-2,-1,0$. There are four, so $$ P(6\le W\le14)=\frac49\approx0.4444. $$ The transformed support values in the interval are $7,9,11,13$, also four values, confirming the answer.

Common mistake avoided: Dividing an inequality by $-2$ without reversing its direction can select the wrong original values. Listing the mapped support provides an independent check.

Common Errors and Useful Checks

Support transformations are simple, but several small errors can change an answer substantially. Before finalizing a calculation, check the support and its probability mass function against the original model.

Counting the endpoints incorrectly. The support from $m$ through $n$ contains $n-m+1$ values, not $n-m$. The difference $n-m$ counts the gaps between endpoints.
Filling in the gaps after scaling. If $b=3$, successive transformed support values differ by 3. Integers between them have probability zero.
Sorting endpoints without considering the sign. A negative $b$ reverses the order. The safest check is to transform both endpoints and identify the smaller and larger results.
Changing the probability denominator. A one-to-one transformation preserves the number of possible outcomes. The denominator remains $N$, not the number of integers between the transformed endpoints.
Using endpoint bounds as if they described every support value. Bounds identify the smallest and largest values, but for $|b|>1$, they do not describe the gaps. State the support as a set or as a sequence with its step size.

For an event involving $Y$, there are two consistent approaches. List the transformed support and count the values in the event, or solve the event inequality in terms of $X$ and count the original support values that satisfy it. When $b<0$, the inequality method requires reversing directions when dividing by $b$. Agreement between the two methods is a useful arithmetic check.

Key takeaway: Under $Y=a+bX$ with $b\ne0$, each original support value maps to exactly one transformed value. The transformed variable is uniform on the mapped support, with probability $1/(n-m+1)$ at each point. A shift moves the support; a scale changes its spacing; a negative scale also reverses its order.

Check Your Understanding

For each problem, identify the transformed support before calculating probabilities. Each model states that all values in the original support are equally likely.

$X$ is uniform on $\{2,3,4,5,6\}$, and $Y=X+8$. Give the support of $Y$ and $P(Y=12)$. (Answer: support $\{10,11,12,13,14\}$; $P(Y=12)=1/5$.)
$X$ is uniform on $\{1,2,\ldots,8\}$, and $Z=3X-2$. Give the support of $Z$ and $P(7\le Z\le16)$. (Answer: support $\{1,4,7,10,13,16,19,22\}$; probability $4/8=1/2$.)
$X$ is uniform on $\{-2,-1,0,1,2,3\}$, and $T=5-2X$. Give the smallest and largest support values, and calculate $P(T>6)$. (Answer: smallest $T=-1$, largest $T=9$; $P(T>6)=2/6=1/3$.)
$X$ is uniform on the integers from 3 through 11, and $R=4X+1$. Calculate $P(21\le R\le37)$ and state why $R=22$ has probability zero. (Answer: $P(21\le R\le37)=5/9$; $R=22$ is not in the support, whose values advance by 4.)
$X$ is uniform on $\{0,1,\ldots,9\}$, and $Q=18-3X$. Find $P(6\le Q\le15)$, showing which original support values produce the event. (Answer: $X=1,2,3,4$; probability $4/10=0.4000$.)

What You'll Learn