Sum of density functions

Question

Consider four pdf $f_1(x), \ldots, f_4(x)$. For any $x$, $f_1(x) \neq \cdots \neq f_4(x)$. Can we prove that $f_1(x) + f_2(x) \neq f_3(x) + f_4(x)$ for some $x$?

Answer 1

That result does not follow from the assumptions, as I will show with a (constructive) counterexample.

Intuition

The idea is that since $f_1+f_2=f_3+f_4$ are sums of pdfs, we can start with that sum (which must be twice some pdf $f$) and try to split it into sums of functions that (i) are themselves pdfs and (ii) never agree at any argument $x.$ The validity of the conclusion comes down to whether this partitioning is ever possible.

The Intermediate Value Theorem of elementary Calculus tells us that if the $f_i$ are continuous, then the necessarily continuous differences $f_i-f_j$ must have zeros: that is, there will be at least one number $x$ at which they agree. Thus, in any counterexample at least three of the $f_i$ must have discontinuities. This leads us to contemplate where these discontinuities occur. In the following construction, they will happen at the boundary of a region $\mathcal A.$

---

Analysis and Notation

Let $f$ be a probability density function (pdf) that is nonzero everywhere. An example is the PDF of any Normal distribution.

Let $\mathcal A$ be any measurable set in the support of $f$ with probability $1/2.$ That is,

$$\frac{1}{2} = \int_{\mathcal A} f(x)\,\mathrm dx.$$

Such sets always exist; for instance, you can take $\mathcal A = (-\infty, m]$ where $m$ is the median of $f.$

As is conventional, let $\mathcal I$ designate an indicator function, so that $\mathcal I_{\mathcal A}(x) = 1$ when $x\in\mathcal A$ and otherwise $\mathcal I_{\mathcal A}(x) = 0.$

I will construct counterexamples determined by a parameter $0\lt p \lt 1$ by defining

$$\phi_p(x) = 2f(x) \left[p \mathcal I_{\mathcal A}(x) + (1-p)\left(1 - \mathcal I_{\mathcal A}(x)\right)\right].$$

(These functions rescale the values of $f$ by $2p$ on the set $\mathcal A$ and by $2(1-p)$ on its complement.)

Every $\phi_p$ is a pdf because (a) its values are non-negative and (b) it integrates to unity:

$$\begin{aligned} \int \phi_p(x)\,\mathrm dx &= 2p\int_{\mathcal A} f(x)\,\mathrm dx + 2(1-p)\left(1 - \int_{\mathcal A} f(x)\,\mathrm dx\right) \\&= 2p\left(\frac{1}{2}\right) + 2(1-p)\left(\frac{1}{2}\right) \\&= 1.\end{aligned}$$

Notice, too, the relationship between each $\phi_p$ and $\phi_{1-p}:$

$$\phi_p(x) + \phi_{1-p}(x) = 2f(x) \left[(p + (1-p)) \mathcal I_{\mathcal A}(x) + ((1-p) + 1-(1-p))\left(1 - \mathcal I_{\mathcal A}(x)\right)\right] = 2f(x)$$

for all $x.$ That is, $\phi_p$ and $\phi_{1-p}$ can serve as possible values of $f_1$ and $f_2$ or as possible values of $f_3$ and $f_4.$

Here's the punch line:

When $p\ne q$ (and both values lie between $0$ and $1$), $\phi_p(x)\ne \phi_q(x)$ for all $x.$

Proof.

Recall $f(x)\ne 0$ for any $x.$ When $x\in\mathcal A,$ $\phi_p(x) = 2p f(x) \ne 2q f(x) = \phi_q(x).$ Otherwise, when $x\notin \mathcal A,$ $\phi_p(x) = 2(1-p)f(x)\ne2(1-q)f(x)=\phi_q(x).$ That covers all the possibilities, QED.

In this schematic (but accurate) figure, $\mathcal A$ is the set of $x$ subtended by the gray region.

Counterexample

Let $p$ and $q$ be two numbers between $0$ and $1,$ but (a) neither equals $1/2,$ (b) $p\ne q,$ and (c) $p\ne 1-q.$ Observe that

$$\phi_p + \phi_{1-p} = 2f = \phi_q + \phi_{1-q},$$

yet since $p,1-p,q$ and $1-q$ are all distinct, all four of the $\phi_*$ in this expression never agree on any $x.$

Answer 2

If you are willing to consider the terminology "pdf" or "density" in its advance (i.e., measure theoretic) sense, more trivial counterexamples (compared with @whuber's great answer) can be easily constructed.

By "advance sense", I meant that given a probability space $(\Omega, \mathscr{F}, P)$, any non-negative measurable function $f: \Omega \to \mathbb{R}$ satisfying \begin{align*} P(A) = \int_A fd\mu, \; A \in \mathscr{F} \end{align*} is defined as a density of the probability measure $P$ with respect to another measure $\mu$ (which dominates $P$). For more discussions on this topic, check this Wolfram MathWorld link and our own Cross Validated discussion.

If you accepted this view, let us set $\Omega = \{0, 1\}$, $\mathscr{F} = 2^\Omega$, $\mu$ is the counting measure, and \begin{align*} f_i(0) = p_i, f_i(1) = 1 - p_i, \quad i = 1, 2, 3, 4. \tag{1}\label{1} \end{align*} It then follows that \begin{align*} & f_1(0) + f_2(0) = p_1 + p_2, f_3(0) + f_4(0) = p_3 + p_4. \end{align*}

It is clear now that for any distinct $0 < p_1, \ldots, p_4 < 1$ such that $p_1 + p_2 = p_3 + p_4$, $\eqref{1}$ can serve as a counterexample. For example, $p_1 = 0.1, p_2 = 0.8, p_3 = 0.3, p_4 = 0.6$.