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!!


1 Answer 1


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" $y=2(m-m^+)$

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

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


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).


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