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

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

• Are you sure there aren't typographical errors in your definition of $\psi$? After all, except at $x=0$ your definition implies $\psi(x) \ge \min(p, 1-p) \gt 0,$ whence (assuming $0 \lt p\lt 1$ and $X$ is not just a constant), $$E(\psi(X-\theta)) \ge \min(p,1-p)(1-\Pr(X=\theta)) \gt 0,$$ whence it is never the case that $E(\psi(X-\theta))=0.$ One way to ensure the result would be to set $\psi=p$ for positive values and $\psi=p-1$ for negative values, as illustrated at stats.stackexchange.com/questions/102986/….
– whuber
Commented Oct 17, 2018 at 21:07
• You are right, I had switched the two terms. Now it should be correct Commented Oct 17, 2018 at 21:14
• That's better: now please fix the typographical errors in the remaining mathematical expressions! That might get you most of the way to the finish.
– whuber
Commented Oct 17, 2018 at 21:17
• Sorry, I fixed the integrals as well. But I'm still lost. How should approach this? Commented Oct 17, 2018 at 21:28
• Write down $\Pr(X\le \theta)$ and $\Pr(X\gt \theta)$ in terms of $f.$
– whuber
Commented Oct 17, 2018 at 21:31

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

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

The second is established as:

\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}

• Thanks! Can you clarify he first equation that you have above? Commented Oct 17, 2018 at 22:15
• The symbol $\mathbb{I}$ is referring here to the indicator function which takes on a value of one if the argument statement is true, and a value of zero if it is false. So you will see that this expression of the function is equivalent to your case-based expression.
– Ben
Commented Oct 17, 2018 at 22:19
• Got it! Thanks. So, from the last equation I get that $P(X<\theta)<p$ but I can't get the $P(X \leq \theta)$. I try simply adding $P(X = 0)$ to get it after rearranging the equation on both sides but it doesn't help.. Any hints? Commented Oct 17, 2018 at 22:30
• This part requires a bit of algebra. Can you show that $b/(1-a) \leqslant b+a$ when $a+b \leqslant 1$?
– Ben
Commented Oct 17, 2018 at 22:32
• No, I just can't get it... Commented Oct 17, 2018 at 23:12

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.

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.

• A good answer (+1), but I think there is presently an error. In the case where $\mathbb{P}(X=\theta) = 1$ and $p>1$, the antecedent conditions for your theorem statement are met, but you get zero expectation. I recommend actually going through the second case explicitly to deal with this. (You actually need an exclusion on this case in the theorem.)
– Ben
Commented Oct 18, 2018 at 22:55
• Thank you, @Ben, for taking such a close look. That nicety does need to be addressed in the second case (which I have left to the reader) and evidently I was too glib in implying nothing goes wrong then. In the portion I have explicitly presented, because it is never the case that $\Pr(X\lt\theta) \gt p \gt 1,$ I don't see any problem. Perhaps I'm overlooking something obvious.
– whuber
Commented Oct 19, 2018 at 11:45
• I think this is just a case where the OP needs to restate the result to rule out that special case. That case is the only one that causes problems, so it is just a matter of ruling it out in the theorem. In any case, I really like your approach to the proof (i.e., demonstrating the contrapositive).
– Ben
Commented Oct 20, 2018 at 2:04