# Conditional expectation of the positive part of y

I'm stuck with the following problem...

Assume that $\mathbb{E}( y \vert x) = a + bx,$ where a and b are parameters. Define $z := max(0, y)$. What are you able to conclude about the relationship between $\mathbb{E}( y \vert x)$ and $\mathbb{E}( z \vert x)$ ?

I tried the following:

Attempt1: Decompose the positive and negative parts of y

$y:= y^{+} - y^{-}$, where $y^{+}=z, y^{-}=-min(0, y)$ are positive and negative parts of y, respectively.

$\mathbb{E}( y \vert x) = \mathbb{E}( y^{+} \vert x) - \mathbb{E}( y^{-} \vert x)$

$= \mathbb{E}( z \vert x) + \mathbb{E}( -y^{-} \vert x)$

$= \mathbb{E}( z \vert x) + \mathbb{E}[ min(0, y) \vert x]$

For the 1st attempt, I tried to find distribution of min(0, y) with cdf(distribution fcn technique) but no assumption is given in the problem, so I failed....

Attempt2: Use CEF-Decomp property (Conditional Expectation Function)

According to Angrist-Pischke, MHE Thm 3.11,

$Y = \mathbb{E}[ Y \vert X ] + \epsilon$, where $\mathbb{E}[\epsilon \vert X] = 0$

$\mathbb{E}( y \vert x) = a + bx \Rightarrow y = a + bx + \epsilon$ $\mathbb{E}( z \vert x) = \mathbb{E}( z \vert x, y \geq 0)P(y \geq 0) + \mathbb{E}( z \vert x, y < 0)P(y < 0)$ $= (a+bx)P(y \geq 0)$

since $\mathbb{E}( z \vert x, y < 0) = 0$

For the 2nd attempt, I failed to find a probability that Y is greater than or equal to zero... Someone told me that this problem is related to order statistics, median or quantile regression. Please help! Thanks in advance!!

The 1st attempt didn't really fail. It's just the best you can do.

The problem does not really depend on $x$. Neither does the linear stuff matter. You could simply ask : I know $E(y)$ what do I know about $E(y^+)$ ? The problem given $x$ is just a special case : same reasoning for each value of $x$.

What you concluded is that $E(y^+)\geq \max(0,E(y))$. The question is : is it possible to say something else ? The answer is no. Let's prove it.

Proof :

Let $m$ and $m^+$ be any two numbers such as $m^+\geq\max(0,m)$. Let's prove that there is some random variable $y$ such as :

• $E(y)=m$
• $E(y^+)=m^+$

Use coin flipping : value "tails" or "heads" have probability $\frac{1}{2}$ each. Define $y$ as :

• if "tails" $y=2m^+$
• if "heads" $y=2(m-m^+)$

$E(y)=\frac{1}{2}2m^+ + \frac{1}{2}2(m-m^+)=m$

$E(y^+)=\frac{1}{2}2m^+ + \frac{1}{2}0=m^+$

Q.E.D

This proof uses a binary variable. But continuous distributions can be used just the same.

As a special case, if $E(y|x)=a+bx$, then $E(y^+|x)\geq\max(0,a+bx)$. That's all you can say. Any stronger result needs to assume something about the distribution of $y$ (given x).