One-to-One Transformations of Discrete Variables

Relabeling the Support

A discrete random variable assigns probability to individual support points. If we apply a one-to-one function to those points, the resulting variable is still discrete: its possible values are the image of the original support, and each image point inherits the probability of its unique preimage. The key task is to keep track of both the new support and the probability attached to each of its points.

This is a point-mass calculation, not a density calculation. There is no Jacobian factor: a transformed point corresponds to one original point, and its probability is unchanged. The map need only be one-to-one on the support that \(X\) actually takes; it does not have to be one-to-one everywhere on the real line.

Definition: Let \(X\) be a discrete random variable with countable support \(S=\{x:P(X=x)>0\}\) and probability mass function \(f_X(x)=P(X=x)\). Let \(T:S\to\mathbb{R}\) be injective, meaning that \(T(x_1)=T(x_2)\) for \(x_1,x_2\in S\) implies \(x_1=x_2\). For \(Y=T(X)\), the image support is \(T(S)=\{T(x):x\in S\}\).

Because \(T\) is injective on \(S\), every \(y\in T(S)\) comes from exactly one point \(x\in S\). We can therefore write \(x=T^{-1}(y)\), where \(T^{-1}\) here means the inverse defined on the image support. This inverse need not be defined on all real numbers.

The Probability Mass Function After Transformation

Theorem: Under the definition above, the probability mass function of \(Y=T(X)\) is

f_Y(y)= \begin{cases} f_X\!\left(T^{-1}(y)\right), & y\in T(S),\\ 0, & y\notin T(S). \end{cases}

In particular, the transformation changes the location of the support points but does not change the probability attached to each point. To obtain a probability of an event, add the masses of all transformed support points in that event. Equivalently, for any set \(A\) of real numbers,

P(Y\in A)=\sum_{\substack{x\in S\\T(x)\in A}} f_X(x).

The sum is over the original support points whose images lie in \(A\). It is valid for any set \(A\), since \(Y\) has countable support. In particular, the cumulative distribution function can be calculated as

F_Y(t)=P(Y\le t)=\sum_{\substack{x\in S\\T(x)\le t}}f_X(x).

This expression does not require \(T\) to be increasing or decreasing. For a non-monotone map, the original points whose images are at most \(t\) may not form a simple interval, so listing or describing those points is often the clearest approach.

Worked Transformations

Worked Example: Shifting a Geometric Random Variable

Let \(X\) have a geometric distribution with success probability \(1/3\), supported on the positive integers: \(f_X(k)=\frac{1}{3}(\frac{2}{3})^{k-1}\) for \(k=1,2,\ldots\). Define \(Y=X+2\). Find the support and probability mass function of \(Y\), and calculate \(P(Y\le 4)\).

Support. The map \(T(k)=k+2\) is injective, and it sends \(1,2,\ldots\) to \(3,4,\ldots\). Thus \(S_Y=\{3,4,\ldots\}\). For any \(j\) in this support, the unique preimage is \(k=j-2\), so

f_Y(j)=f_X(j-2) =\frac{1}{3}\left(\frac{2}{3}\right)^{(j-2)-1} =\frac{1}{3}\left(\frac{2}{3}\right)^{j-3}, \qquad j=3,4,\ldots.

For example, \(P(Y=3)=f_X(1)=1/3\), and \(P(Y=4)=f_X(2)=2/9\). The masses sum to one:

\sum_{j=3}^{\infty} f_Y(j) =\frac{1}{3}\sum_{r=0}^{\infty}\left(\frac{2}{3}\right)^r =\frac{1}{3}\cdot\frac{1}{1-2/3}=1.

Finally, \(Y\le4\) corresponds to \(X+2\le4\), or \(X\le2\). Therefore \(P(Y\le4)=P(X=1)+P(X=2)=1/3+2/9=5/9\).

Worked Example: Rescaling a Binomial Random Variable

Suppose \(X\sim\operatorname{Binomial}(3,1/2)\), so \(S_X=\{0,1,2,3\}\) and the respective probabilities are \(1/8,3/8,3/8,1/8\). Let \(Y=2X+1\). Find its support and mass function, then calculate \(P(Y\le4)\).

The map \(T(x)=2x+1\) is injective. Applying it to each support point gives \(T(0)=1\), \(T(1)=3\), \(T(2)=5\), and \(T(3)=7\). Each image keeps the mass of its unique preimage:

\begin{array}{c|cccc} y & 1 & 3 & 5 & 7\\ \hline f_Y(y) & 1/8 & 3/8 & 3/8 & 1/8 \end{array}

The masses still sum to \(1/8+3/8+3/8+1/8=1\). For the requested probability, the image support points no greater than \(4\) are \(1\) and \(3\), so \(P(Y\le4)=1/8+3/8=1/2\). Equivalently, \(2X+1\le4\) requires \(X\le1\), giving \(P(X=0)+P(X=1)=1/2\).

Worked Example: Squaring a Poisson Random Variable

Let \(X\sim\operatorname{Poisson}(2)\), with \(f_X(k)=e^{-2}2^k/k!\) for \(k=0,1,2,\ldots\). Define \(Y=X^2\). Although squaring is not injective on all real numbers, it is injective on the nonnegative integer support of \(X\). Find the support and mass function of \(Y\), and calculate \(P(Y\le4)\).

The image support is \(S_Y=\{0,1,4,9,16,\ldots\}\), the set of squares of nonnegative integers. Each \(m^2\) has the unique preimage \(m\) in the support, so

f_Y(m^2)=f_X(m)=e^{-2}\frac{2^m}{m!}, \qquad m=0,1,2,\ldots,

and \(f_Y(y)=0\) at other real values. The first four image-point masses are \(f_Y(0)=e^{-2}\), \(f_Y(1)=2e^{-2}\), \(f_Y(4)=2e^{-2}\), and \(f_Y(9)=\frac{4}{3}e^{-2}\). They sum over the full support to

\sum_{m=0}^{\infty} f_Y(m^2) =e^{-2}\sum_{m=0}^{\infty}\frac{2^m}{m!}=1,

as required. The condition \(Y\le4\) is equivalent here to \(X^2\le4\), which, since \(X\ge0\), is equivalent to \(X\in\{0,1,2\}\). Hence \(P(Y\le4)=e^{-2}(1+2+2)=5e^{-2}\), approximately \(0.6767\).

Expectations of Transformed Variables

The same point-mass rule gives a direct way to calculate expectations. It is especially useful when the desired quantity is a function of the transformed variable. Let \(h\) be a measurable real-valued function on the image support. If \(h(Y)\) is nonnegative, its expectation may be infinite; if it takes both positive and negative values, assume \(E[|h(Y)|]<\infty\).

Theorem: Under either of those conditions,

E[h(Y)] =\sum_{y\in T(S)}h(y)f_Y(y) =\sum_{x\in S}h(T(x))f_X(x).

Proof. By the definition of expectation for a discrete random variable, \(E[h(Y)]=\sum_{y\in T(S)}h(y)f_Y(y)\), with the sum interpreted in the extended nonnegative sense when \(h(Y)\ge0\). Substitute the transformed mass function. Each \(y\in T(S)\) has exactly one preimage \(x\in S\), so re-indexing the sum by that preimage gives

\sum_{y\in T(S)}h(y)f_Y(y) =\sum_{x\in S}h(T(x))f_X(x).

For nonnegative \(h\), re-indexing a countable sum of nonnegative terms is valid even if its value is infinite. If \(E[|h(Y)|]<\infty\), the corresponding series is absolutely convergent, so the re-indexing is also valid. This proves the identity. \(\square\)

Worked Example: An Expectation After a Nonlinear Transformation

For the Poisson example above, calculate \(E[Y]\) using \(Y=X^2\). The expectation-transfer formula with \(h(y)=y\) gives

E[Y] =\sum_{k=0}^{\infty} k^2 e^{-2}\frac{2^k}{k!}.

The \(k=0\) term is zero. For \(k\ge1\), write \(k^2=k(k-1)+k\), and split the sum:

E[Y] =e^{-2}\sum_{k=2}^{\infty}k(k-1)\frac{2^k}{k!} +e^{-2}\sum_{k=1}^{\infty}k\frac{2^k}{k!}.

In the first sum, \(k(k-1)/k!=1/(k-2)!\); in the second, \(k/k!=1/(k-1)!\). Thus

E[Y] =e^{-2}\left(4\sum_{r=0}^{\infty}\frac{2^r}{r!} +2\sum_{r=0}^{\infty}\frac{2^r}{r!}\right) =e^{-2}(4e^2+2e^2)=6.

The calculation uses the transformed expectation formula and the exponential series. It does not require finding a density for \(Y\); indeed, \(Y\) has mass only at nonnegative perfect squares.

Check Injectivity on the Support

The condition that matters is injectivity on the values \(X\) can actually take. A function may fail to be one-to-one on the real line but still be one-to-one on a particular discrete support, as squaring is on the nonnegative integers. Conversely, a formula that appears simple does not guarantee injectivity on the relevant support; it must be checked there.

A common mistake is to use the unique-preimage formula when two support points have the same image. For instance, if \(X\) is uniform on \(\{-1,1\}\) and \(T(x)=x^2\), both original points map to \(1\). The event \(\{T(X)=1\}\) includes both possibilities, so its probability is \(1/2+1/2=1\), not the mass of just one preimage. In general, if injectivity fails, probabilities at a shared image must be added over all its preimages; the theorem's single-preimage formula does not apply.

For an injective transformation, a reliable calculation has three parts: map every original support point, keep the probability paired with that point, and then calculate event probabilities by summing the masses at the relevant image points. This avoids treating a discrete support as though it were a continuous interval or confusing point probabilities with density values.

Check Your Understanding

Use the support and mass-function rules to answer these questions.

If \(X\) has support \(\{0,1,2\}\) with probabilities \(0.2,0.5,0.3\), what are the support points and masses of \(Y=3-X\)?
For a discrete \(X\), why is it enough for \(T\) to be injective on the support of \(X\), rather than on all real numbers?
If \(Y=T(X)\) and \(T\) is injective on the support, how would you express \(P(Y\le t)\) as a sum over the original support?
Let \(X\) be uniform on \(\{-2,2\}\) and \(Y=X^2\). Why can the unique-preimage formula not be used, and what is \(P(Y=4)\)?
State a condition under which \(E[h(Y)]=\sum_{x\in S}h(T(x))f_X(x)\) is valid when \(h\) can take both positive and negative values.