Show that if $E\psi(x-\theta)= 0 $ then $P(X< \theta) \leq p \leq P(X \leq \theta)$

Question

Define $$\psi(x)=\begin{cases} 1-p & x < 0 \\ 0 & x=0 \\ -p & x> 0 \end{cases}$$. I have to show that if $$E\psi(x-\theta)= 0 $$ then $$P(X< \theta) \leq p \leq P(X \leq \theta)$$.

Any hints will be appreciated. What I have done so far is to state

$$E\psi(x-\theta)= \int_{-\infty}^\theta{(1-p)f(x)dx} - p \int_\theta^{\infty} {f(x)dx} = 0$$

Then what I get is that

$$\int_{-\infty}^\theta{(1-p)f(x)dx} = p \int_\theta^{\infty} {f(x)dx}$$

But this doesn't help much

Answer 1

Your method assumes that $X$ is a continuous random variable, which is not stated as a condition of the problem. It is possible to get the result in a more general case, so long as there is some non-zero probability that $X \neq \theta$. From your stated form for $\psi$ you have:

$$\psi(X-\theta) = \mathbb{I}(X < \theta) - p \cdot \mathbb{I}(X \neq \theta).$$

Hence, taking the expected value you get the function:

$$\begin{equation} \begin{aligned} \mu(\theta) \equiv \mathbb{E}(\psi(X-\theta)) &= \mathbb{P}(X < \theta) - p \cdot \mathbb{P}(X \neq \theta). \\[6pt] \end{aligned} \end{equation}$$

Now, for the case where $\mathbb{P}(X \neq \theta)=0$ you have $\mu(\theta)=0$ for any $p \in \mathbb{R}$, so the implication you want to prove is not present in that case. For the case where $\mathbb{P}(X \neq \theta)>0$ the expected value condition $\mu(\theta)=0$ implies that:

$$p = \frac{\mathbb{P}(X < \theta)}{1-\mathbb{P}(X = \theta)}.$$

With a bit of algebra you can obtain the required inequalities. The first is established by:

$$\begin{equation} \begin{aligned} p = \frac{\mathbb{P}(X < \theta)}{1-\mathbb{P}(X = \theta)} \geqslant\mathbb{P}(X < \theta). \end{aligned} \end{equation}$$

The second is established as:

$$\begin{equation} \begin{aligned} p &= \frac{\mathbb{P}(X < \theta)}{1-\mathbb{P}(X = \theta)} \\[6pt] &\leqslant \frac{\mathbb{P}(X < \theta) + \mathbb{P}(X = \theta) \cdot \mathbb{P}(X > \theta)}{1-\mathbb{P}(X = \theta)} \\[6pt] &= \frac{\mathbb{P}(X < \theta) + \mathbb{P}(X = \theta) \cdot (1-\mathbb{P}(X < \theta)-\mathbb{P}(X = \theta))}{1-\mathbb{P}(X = \theta)} \\[6pt] &= \frac{\mathbb{P}(X < \theta) + \mathbb{P}(X = \theta) - \mathbb{P}(X = \theta)\mathbb{P}(X < \theta)-\mathbb{P}(X = \theta)^2}{1-\mathbb{P}(X = \theta)} \\[6pt] &= \frac{(\mathbb{P}(X < \theta) + \mathbb{P}(X = \theta)) \cdot (1 - \mathbb{P}(X = \theta))}{1-\mathbb{P}(X = \theta)} \\[6pt] &= \mathbb{P}(X < \theta) + \mathbb{P}(X = \theta) \\[8pt] &= \mathbb{P}(X \leqslant \theta). \\[6pt] \end{aligned} \end{equation}$$

Answer 2

There are many ways to approach this problem.

The point of the following is to take you through a process of analyzing the question, performing the requisite calculations as simply and easily as possible, developing a strategy to carry out the proof, and applying that strategy. A section of concluding remarks highlights what has been achieved by this method.

Analysis

Recall that a random variable $X$ assigns a number to each element $\omega$ of a sample space $\Omega.$

The expression $Y=\psi(X-\theta)$ is another way to assign numbers to elements of $\Omega;$ namely, for each $\omega,$ compute $\psi(X(\omega)-\theta).$

Notice that the resulting number is one of at most three values: $-p, 0,$ and $1-p.$ As you will see, we will be able to compute the probabilities of any subset of these values. This makes $Y$ a random variable, too: and it's a discrete one. This simplifies the calculations we might need to do.

At this point it is evident we need to accomplish two things: (1) we need to compute an expectation and (2) we will need to manipulate inequalities algebraically. Let's take these in turn.

Preliminary calculations

The expectation of $Y$ can be found from the very definition: multiply its values by its probabilities. Let's tabulate them:

1. For $X \lt \theta,$ $Y=\psi(X-\theta) = 1-p.$ This has probability $\Pr(X\lt \theta).$

2. For $X=\theta,$ $Y=\psi(0) = 0.$ This has probability $\Pr(X=\theta).$

3. For $X \gt \theta,$ $Y = -p.$ This has probability $\Pr(X \gt \theta).$

The expectation of $Y$ is the sum of its values times their probabilities:

$$\mathbb{E}(\psi(X-\theta)) = \mathbb{E}[Y] = (1-p)\Pr(X\lt\theta) + 0\Pr(X=\theta) + (-p)\Pr(X \gt \theta).\tag{1}$$

Strategy for the proof

The simplest way to carry out the required demonstration is to show its contrapositive: that is,

if $\Pr(X \lt \theta)\gt p$ or $\Pr(X \le \theta) \lt p,$ we need to conclude that $\mathbb{E}(\psi(X-\theta))$ is not zero.

In the first case where $\Pr(X\lt \theta)\gt p,$ the additivity of mutually exclusive events and the axiom of unit measure--two of the axioms of probability--guarantee that

$$\eqalign{ \Pr(X \gt \theta) &= 1 - (\Pr(X\lt\theta) + \Pr(X=\theta)) \\ &\le 1 - \Pr(X\lt \theta)\\ & \lt 1-p. }$$

Substituting these two inequalities into $(1)$ gives

$$\eqalign{ \mathbb{E}(\psi(X-\theta)) &= (1-p)\Pr(X\lt\theta) -p\Pr(X \gt \theta) \\ &\gt (1-p)p - p(1-p) = 0, }$$

proving the expectation cannot be zero. The demonstration in the second case parallels this one, QED.

Comments

Because this proof is completely elementary--it relies only on the definition of expectation and axioms of probability--it reveals how little needs to be assumed and how general the result is:

1. It is not necessary to assume $0\lt p\lt 1:$ the assertion that was proved is true for all $p.$

2. This demonstration works even when $p=0$ or $1-p=0$ (whereas other attempts might fail for these values, in case they appear in the denominators of any fractions).

3. It was not necessary to assume $X$ has a density $f.$

4. No higher mathematical concepts, such as integration, were needed.

5. Indeed, we did not even have to use a distribution function for $X$: we worked directly with the relevant probabilities.