<p>To be fair to the author, the quote says that "it is unlikely", not that "it is impossible".  </p>

<p>But apart from that, the link between "same estimates" and "linearity of the conditional relations" is misleading. </p>

<p>Example: assume that for an unconditional distribution, the median $m$ is a linear function of the mean $\mu$ : </p>

<p>$$m = c\mu$$</p>

<p>(this is for example approximately the case <a href="https://en.wikipedia.org/wiki/Gamma_distribution#Median_calculation" rel="nofollow noreferrer">for the Gamma distribution</a>).</p>

<p>Assume that we model the conditional distribution of $Y$ given $X$ as a Gamma distribution, with </p>

<p>$$E(Y\mid X) = bX$$</p>

<p>It then follows that the conditional median will be</p>

<p>$$m(Y\mid X) = c(bX) = (cb)X = \delta X, \delta \neq b$$</p>

<p>Here, the conditional median can be modeled as a linear function of the conditioning variables also, but we expect that the estimates we will obtain will be different.</p>

CC BY-SA 3.0
2018-04-28T11:37:32.737Z
2018-04-28T11:37:32.737Z
Alecos Papadopoulos
<p>In Alecos Papadopoulos's answer, both the conditional mean and the conditional median are linear in $X$. In the following example, the conditional mean is linear in $X$ while the conditional median is quadratic in $X$. This example is a constructed toy but it touches the heart of the issue.</p>

<p><strong>Example:</strong> Suppose that $y = \beta_0 + \beta_1 x + u$ and $u = x^2 (w - 1)$, where $w \sim \chi_1^2$ and $x$ and $w$ are mutually independent. Then, as the R command <code>qchisq(0.5,1)</code> gives, $med(w) \doteq 0.455$ so that $med(u|x) = x^2 (-0.545)$, while $E(u|x) = 0$.  We thus have
\begin{equation}
  E(y|x) = \beta_0 + \beta_1 x, ~~\text{ while }~~ med(y|x) = \beta_0 + \beta_1 x - 0.545 x^2.
\end{equation}</p>

<p><strong>Note:</strong> Asymmetry of the error distribution and the dependence of $u$ on $x$ are both important. If the error distribution is symmetric, then the conditional mean function and the conditional median function coincide. Also, if the distribution of $u$ is independent of $x$, then $med(u|x)$ is nonzero but constant and thus $med(y|x)$ is again linear in $x$.</p>

CC BY-SA 3.0
2018-04-29T13:53:01.917Z
2018-04-29T13:53:01.917Z
chan1142
<p>In Wooldridge's book (Page 452), it says</p>

<blockquote>
  <p>When linear absolute deviation (LAD) methods are applied alongside OLS, thre are often reasons to think a priori that OLS and LAD will not produce similar slope estimates. (in fact, it is unlikely that the conditional mean and conditional median are both linear in $x_i$).</p>
</blockquote>

<p>I understand that LAD estimates and OLS estimates are generally different, but the words in parentheses confuse me. Why they cannot be both linear? Could anyone give me an example? </p>

CC BY-SA 3.0
2018-04-28T10:23:45.320Z
For example, the case the error distribution is asymmetric and heteroskedastic.
chan1142
CC BY-SA 3.0
2018-04-28T17:45:20.777Z
@chan1142 Could you explain how error being asymmetrically distributed imply the non-linearity of conditional median in $x_1$?
Chunjing GU
CC BY-SA 3.0
2018-04-29T13:55:25.350Z
Chunjing: It's too long for a comment. Please see my answer below.
chan1142
<econometrics><regression>
conditional mean and conditional median
Chunjing GU
# conditional mean and conditional median

(Asked by: username_0 on 2018-04-27)

In Wooldridge's book (Page 452), it says

> When linear absolute deviation (LAD) methods are applied alongside OLS, thre are often reasons to think a priori that OLS and LAD will not produce similar slope estimates. (in fact, it is unlikely that the conditional mean and conditional median are both linear in $x\_i$).

I understand that LAD estimates and OLS estimates are generally different, but the words in parentheses confuse me. Why they cannot be both linear? Could anyone give me an example?

**Tags:** econometrics, regression
> **Comments:**
> **username_1**: For example, the case the error distribution is asymmetric and heteroskedastic.  
> **username_0**: @chan1142 Could you explain how error being asymmetrically distributed imply the non-linearity of conditional median in $x_1$?  
> **username_1**: Chunjing: It's too long for a comment. Please see my answer below.  

---

## 2 answers

### Answer by username_1 on 2018-04-29. Score: 3 | Accepted answer

In Alecos Papadopoulos's answer, both the conditional mean and the conditional median are linear in $X$. In the following example, the conditional mean is linear in $X$ while the conditional median is quadratic in $X$. This example is a constructed toy but it touches the heart of the issue.

**Example:** Suppose that $y = \beta\_0 + \beta\_1 x + u$ and $u = x^2 (w - 1)$, where $w \sim \chi\_1^2$ and $x$ and $w$ are mutually independent. Then, as the R command `qchisq(0.5,1)` gives, $med(w) \doteq 0.455$ so that $med(u|x) = x^2 (-0.545)$, while $E(u|x) = 0$. We thus have
\begin{equation}
E(y|x) = \beta\_0 + \beta\_1 x, ~~\text{ while }~~ med(y|x) = \beta\_0 + \beta\_1 x - 0.545 x^2.
\end{equation}

**Note:** Asymmetry of the error distribution and the dependence of $u$ on $x$ are both important. If the error distribution is symmetric, then the conditional mean function and the conditional median function coincide. Also, if the distribution of $u$ is independent of $x$, then $med(u|x)$ is nonzero but constant and thus $med(y|x)$ is again linear in $x$.


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### Answer by username_2 on 2018-04-28. Score: 2

To be fair to the author, the quote says that "it is unlikely", not that "it is impossible".

But apart from that, the link between "same estimates" and "linearity of the conditional relations" is misleading.

Example: assume that for an unconditional distribution, the median $m$ is a linear function of the mean $\mu$ :

$$m = c\mu$$

(this is for example approximately the case [for the Gamma distribution](https://en.wikipedia.org/wiki/Gamma_distribution#Median_calculation)).

Assume that we model the conditional distribution of $Y$ given $X$ as a Gamma distribution, with

$$E(Y\mid X) = bX$$

It then follows that the conditional median will be

$$m(Y\mid X) = c(bX) = (cb)X = \delta X, \delta \neq b$$

Here, the conditional median can be modeled as a linear function of the conditioning variables also, but we expect that the estimates we will obtain will be different.


---

