Transforming Mixed Distributions

One Transformation, Two Parts of a Distribution

A mixed distribution assigns positive probability to particular points while also spreading probability continuously over intervals. Applying a function to a random variable moves both parts of its law, but they must be tracked separately: point masses remain point masses, whereas the continuous part is transformed by a density change-of-variables calculation.

Let $Y=T(X)$, where $T$ is one-to-one on an interval containing the support of $X$. Injectivity ensures that different input values do not collapse to the same output. Under a suitable differentiability condition, it also lets us use one inverse function to find the density of the transformed continuous component.

Definition: A mixed distribution of the type considered here has a law that can be written as a countable collection of atoms and an absolutely continuous component. Specifically, for a random variable $X$ taking values in an interval $I$,

$$ P(X\in A)=\sum_j p_j\,\mathbf{1}\{x_j\in A\}+\int_{A\cap I}g(x)\,dx $$
for every Borel set $A$, where the distinct points $x_j\in I$ have masses $p_j>0$, $g$ is a nonnegative density on $I$, and $\sum_j p_j+\int_I g(x)\,dx=1$. There is no additional singular continuous component.

The atom at $x_j$ means $P(X=x_j)=p_j$. The density $g$ describes the rest of the probability: a density value $g(x)$ is not itself a point probability. In particular, the integral component assigns probability zero to every singleton.

The Transformation Rule

Assume $T:I\to T(I)$ is strictly monotone and continuously differentiable, with $T'(x)\ne 0$ throughout the interior of $I$. These conditions make $T$ invertible on its image and allow the one-dimensional change-of-variables formula to be used for the continuous component. Write $h=T^{-1}$. For $Y=T(X)$, the transformed atoms and continuous density are as follows.

Theorem: Under the assumptions above, $Y$ has an atom of probability $p_j$ at $T(x_j)$ for each atom $x_j$ of $X$. Its absolutely continuous component has density

$$ g_Y(y)= \begin{cases} \dfrac{g(h(y))}{|T'(h(y))|}, & y\in T(I),\\[6pt] 0, & y\notin T(I). \end{cases} $$
The density formula is understood on the interior of $T(I)$; endpoint values do not affect probabilities.

To see why the formula treats the two parts separately, take any Borel set $B$. The probability that the transformed variable falls in $B$ is

P(Y\in B)=\sum_j p_j\,\mathbf{1}\{T(x_j)\in B\} +\int_I \mathbf{1}\{T(x)\in B\}g(x)\,dx.

The first term maps each atom to its new location. For the second term, the ordinary one-dimensional substitution $y=T(x)$ gives the density formula, with the absolute value accounting for either direction of monotonicity. Thus the transformed law is the sum of the mapped point masses and the transformed density. This proves the stated decomposition. $\square$

The atom probabilities can also be checked directly. Because $T$ is injective, $T(x)=T(x_j)$ holds only when $x=x_j$. Therefore

$$ P(Y=T(x_j))=P(X=x_j)=p_j. $$

The continuous component contributes zero probability to this singleton: its preimage is the singleton $\{x_j\}$, which has probability $\int_{\{x_j\}}g(x)\,dx=0$. Injectivity also ensures that distinct atoms remain at distinct transformed locations. Finally, the transformed density integrates to $\int_I g(x)\,dx$, by the same change of variables. Adding the unchanged atom probabilities shows that the transformed parts still sum to one.

Worked Transformations

Worked Example: An Increasing Transformation With One Atom

Suppose $X$ has probability $1/3$ at $x=2$, and its remaining probability $2/3$ is uniform on $(-1,1)$. Thus its continuous density is $g(x)=1/3$ for $-1<x<1$, and zero elsewhere. Let $Y=\arctan(X)$.

The atom maps to $T(2)=\arctan(2)$. Since $T$ is one-to-one, the transformed atom remains a single atom with probability $1/3$. On the continuous component, the inverse is $h(y)=\tan(y)$, and $T'(x)=1/(1+x^2)$. The image of $(-1,1)$ is $(-\pi/4,\pi/4)$. Hence

g_Y(y)=\frac{1/3}{1/(1+\tan^2 y)} =\frac{1}{3}\sec^2(y), \qquad -\frac{\pi}{4}<y<\frac{\pi}{4}.

The continuous probability is $ (1/3)[\tan(y)]_{-\pi/4}^{\pi/4}=2/3$, as required. The complete transformed law therefore has density $(1/3)\sec^2(y)$ on that interval and an atom of mass $1/3$ at $\arctan(2)$. For a check on the distinction, $P(Y=0)=P(X=0)=0$: the density is positive at zero, but there is no atom there.

Worked Example: A Decreasing Transformation

Let $X$ have probability $1/4$ at $x=2$, with its remaining probability $3/4$ uniformly distributed on $(0,1)$. Set $Y=T(X)$, where $T(x)=-x-x^3$. This function is strictly decreasing because $T'(x)=-1-3x^2<0$.

The atom maps to $T(2)=-10$, retaining probability $1/4$. On the continuous component, $T$ maps $(0,1)$ onto $(-2,0)$. For $-2<y<0$, let $u=h(y)$ be the unique solution in $(0,1)$ of $u+u^3=-y$. The input density on $(0,1)$ is $3/4$, and the absolute derivative is $1+3u^2$. Thus

g_Y(y)=\frac{3}{4(1+3u^2)},\qquad -2<y<0, \quad\text{where }u+u^3=-y.

Although $u$ need not be expressed in a convenient closed form, it is uniquely determined for every $y$ in the image interval. The continuous component has total probability

\int_{-2}^{0}g_Y(y)\,dy =\int_0^1 \frac{3}{4(1+3x^2)}(1+3x^2)\,dx =\frac{3}{4}.

The derivative factor in this check is $|T'(x)|=1+3x^2$; the decreasing direction is handled by the absolute value. Together, the density and the atom at $-10$ account for $3/4+1/4=1$.

Worked Example: An Atom at a Point Also Covered by the Density

Suppose $X$ has probability $1/4$ at zero, and with probability $3/4$ is uniform on $(-1,1)$. The continuous density is therefore $3/8$ on $(-1,1)$. Let $Y=X+X^3$. Since $T'(x)=1+3x^2>0$, this transformation is strictly increasing. The atom maps to zero and keeps its probability $1/4$.

For $y\in(-2,2)$, let $u$ be the unique solution of $u+u^3=y$ in $(-1,1)$. The transformed continuous density is

g_Y(y)=\frac{3}{8(1+3u^2)},\qquad -2<y<2.

In particular, the density is positive at $y=0$, where $u=0$ and $g_Y(0)=3/8$. But this does not add a point probability: the atom at zero has probability $1/4$, while the continuous component assigns probability zero to $\{0\}$. The continuous mass checks as

\int_{-2}^{2}g_Y(y)\,dy =\int_{-1}^{1}\frac{3}{8(1+3x^2)}(1+3x^2)\,dx =\frac{3}{4}.

Thus the transformed law has both an atom at zero and a continuous density whose support includes zero. Their coexistence is valid because a density describes probability over intervals, not probability concentrated at a single point.

Common Mistakes and Why the Distinction Matters

The most common error is to transform the density formula and then treat it as the entire law. That discards every atom. The opposite error is to list the transformed atoms but leave the continuous density unchanged. That generally gives incorrect interval probabilities because the transformation stretches or compresses the continuous part.

Forgetting atom locations or masses. Map each atom $x_j$ to $T(x_j)$, and retain its probability $p_j$.
Using $T'$ without an absolute value. For a decreasing transformation, the derivative is negative, but a density must be nonnegative. Use $|T'|$.
Reading a density value as a point probability. A positive value $g_Y(y)$ does not imply $P(Y=y)>0$. Point masses must be identified separately.
Allowing distinct inputs to merge without adjustment. The rule here relies on injectivity. If a function is not one-to-one, multiple preimages may contribute to the same output, so this single-inverse formula does not apply as stated.

Keeping the two components separate is useful in probability calculations and statistical models: the transformed law may have jumps in its cumulative distribution function at atoms and a continuous change between those jumps. A complete description must preserve both features, rather than forcing the entire law into a density.

Key takeaway: Under an injective differentiable transformation, each atom moves to its image with unchanged probability, while the continuous component receives the usual change-of-variables density. Check both parts and verify that their total probability is one.

Check Your Understanding

For each question, keep the point masses and the continuous component distinct.

If $P(X=1)=0.2$ and $Y=T(X)$ for an injective function $T$, what is the probability of the atom at $T(1)$?
Why can a transformed density be positive at $y_0$ even when $P(Y=y_0)=0$?
For a strictly decreasing differentiable transformation, why does the density formula use $|T'(h(y))|$ rather than $T'(h(y))$?
A mixed law has atom masses summing to $0.3$ and a continuous component with total probability $0.7$. What should the transformed atom masses and density integrate to?
What feature of the transformation guarantees that two distinct atoms of $X$ cannot map to the same transformed location?

Transforming Mixed Distributions

What You'll Learn