---
# System prepended metadata

title: Core August 2020 Metrics
tags: [' group', ' core', metrics]

---

---
tags: metrics, core, group
---

$$
% My definitions 
\def\ve{{\varepsilon}}
\def\dd{{\text{ d}}}
\newcommand{\dif}[2]{\frac{d #1}{d #2}} % for derivatives
\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}} % for partial derivatives
\def\R{\text{R}}
\def\E{\mathbb{E}}
$$

# Q1
Let a random variable $X$ have the following distribution:
$$
P(X=\frac{1}{4})= \frac{1}{4}, P(X=\frac{1}{2})=\frac{1}{2}, P(X=\frac{3}{4})=\frac{1}{4}
$$

Suppose $Y$ is a Bernoulli random variable and the joint distribution of $X$ and $Y$ satisfies the following condition:
$$\E[Y |X] = X^2$$

Calculate:
(a) $\E[Y ]$.
(b) $\E[XY ]$.
(c ) $\E[ \frac{Y}{X}]$.
(d) $Var[Y]$.
(e) $Var[Y|X]$.
(f) $\E[X|Y]$.

## Answer
### a

$$
\E[Y] = \E[\E[Y|X]] = \E[X^2] = \frac{9}{32},
$$

By LIE.

### b
$$
\E[XY] = \E[\E[XY|X]] = \E[X\E[Y|X]] = \E[X^3] =\frac{11}{64} ,
$$
By LIE.

### c
$$
\E[\frac{Y}{X}] = \E[\E[\frac{Y}{X}|X]] = \E[\frac{1}{X}\E[Y|X]] = \E[X] =\frac{1}{2} ,
$$
By LIE.

### d
Since $Y$ is  a Bernoulli random variable, let $p=P(Y=1)$, we know $Var[Y]=p(1-p)$. Moreover,

$$
p=P(Y=1) = \E[Y] = \frac{9}{32} \implies Var[Y]= \frac{9}{32} \cdot \frac{23}{32}
$$

### e

$$
Var[Y|X] = \E[Y^2|X] - \E[Y|X]^2)
$$

Since $Y$ is  a Bernoulli random variable, the outcome of $Y$ is still $\{0,1\}$ conditional on $X$. In other words, $Y|X$ is a Bernoulli random variable with differe parameter $p(x)$. Moreover, the event $Y^2=1$ is  the same event of $Y=1$;  the event $Y^2=0$ is the same event of $Y=0$. Hence,


$$
\E[Y^2|X=x]  = \E[Y|X=x]
$$

Hence,

$$
Var[Y|X] = X^2 - (X^2)^2 = X^2 - X^4.
$$

### f
We can use Bayes's rule.


Let $p(X)$ be the parameter of Bernoulli r.v. $Y|X$.

$$
\E[Y|X=\frac{1}{4}] = \frac{1}{16} \implies p(X=\frac{1}{4})= \frac{1}{16}\\
\E[Y|X=\frac{1}{2}] = \frac{1}{4} \implies p(X=\frac{1}{2})= \frac{1}{4}\\\
\E[Y|X=\frac{3}{4}] = \frac{9}{16} \implies p(X=\frac{3}{4})= \frac{9}{16}\\
$$

We already know $P(Y=1)= \E[Y] = \frac{9}{32}$. Hence, given $Y=1$,

$$
P(X=\frac{1}{4}|Y=1) = \frac{\frac{1}{64}}{\frac{9}{32}} = \frac{1}{18}\\
P(X=\frac{1}{2}|Y=1) = \frac{\frac{8}{64}}{\frac{9}{32}} = \frac{4}{9}\\\
P(X=\frac{3}{4}|Y=1) = \frac{\frac{9}{64}}{\frac{9}{32}} = \frac{1}{2}\\
$$

Hence,

$$
\E[X|Y=1]= \frac{1}{4} \cdot \frac{1}{18} + \frac{1}{2} \cdot \frac{4}{9} + \frac{3}{4} \cdot \frac{1}{2}= \frac{11}{18}
$$

We can also derive $\E[X|Y=0]$.

# Q2
Consider the simple regression model

$$Y_i = \beta_0 + \beta_1 X_i + u_i$$

and let $Z_i$ be a dummy instrumental variable for $X_i$, where $E(u|Z) = 0.$

(a) Show that the IV estimator $\hat{\beta}_1$ can be written as
$$
\hat{\beta}_1  = \frac{\hat{Y}_1-\hat{Y}_0}{\bar{X}_1-\bar{X}_0}.
$$
where $\hat{Y}_0$ and $\hat{X}_0$ are the sample averages of $Y_i$ and $X_i$ over the part of the sample with $Z_i= 0$, and where $\hat{Y}_1$ and $\hat{X}_1$ are the sample averages of $Y_i$ and $X_i$ over the part of the sample with $Z_i = 1$.
(b) Is it a consistent estimator of $\beta_1$? Why?
## Answer
### a
:::info
Derive from the scratch
:::

$$
Z'=
\begin{pmatrix}
1 & \cdots & 1 \\
z_1 & \cdots & z_n
\end{pmatrix},
X=
\begin{pmatrix}
1 & x_1 \\
& \cdots \\
1 & x_n
\end{pmatrix},
Y=
\begin{pmatrix}
y_1 \\
 \cdots \\
y_n
\end{pmatrix}
$$

Hence,

$$
Z'X= \begin{pmatrix}
n & \sum_{i=1}^n x_i \\
\sum_{i=1}^n z_i & \sum_{i=1}^n z_i x_i
\end{pmatrix},
Z'Y= \begin{pmatrix}
\sum_{i=1}^n y_i\\
 \sum_{i=1}^n z_i y_i 
\end{pmatrix} = \begin{pmatrix}
n \E_n[y]\\
 n\E_n[zy]
\end{pmatrix}
$$

Also,

$$
(Z'X)^{-1} = \frac{1}{n \sum_{i=1}^n z_i x_i - \sum_{i=1}^n  x_i  \sum_{i=1}^n z_i} \begin{pmatrix}
\sum_{i=1}^n z_i x_i & -\sum_{i=1}^n x_i \\
-\sum_{i=1}^n z_i & n
\end{pmatrix}\\
= \frac{1}{n^2 \E_n[zx] - n^2 \E_n[x] \E_n[z]}  \begin{pmatrix}
n \E_n[zx] & -n \E_n[x] \\
-n \E_n[z] & n
\end{pmatrix}\\
$$

Thus,

$$
(Z'X)^{-1} Z'Y =  \frac{1}{n \sum_{i=1}^n z_i x_i - \sum_{i=1}^n  x_i}   \begin{pmatrix}
\sum_{i=1}^n z_i x_i \sum_{y_i} - \sum_{i=1}^n x_i \sum_{i=1}^n z_i y_i \\ -\sum_{i=1}^n z_i \sum_{i=1}^n y_i + n \sum_{i=1}^n z_i y_i
\end{pmatrix}\\
= \frac{1}{n^2 \E_n[zx] - n^2 \E_n[x] \E_n[z]}  \begin{pmatrix}
n^2 \E_n[zx] \E_n[y] - n^2 \E_n[x] \E_n[zy]\\
-n^2 \E_n[z] \E_n[y] + n^2 \E_n[zy]
\end{pmatrix}\\
$$

Hence, 

$$
\hat{\beta}_1 = \frac{n^2 \E_n[y] -n^2 \E_n[z] \E_n[zy]}{n^2 \E_n[zx] - n^2 \E_n[x] \E_n[z]} = \frac{ \E_n[zy] - \E_n[z] \E_n[y]}{ \E_n[zx] -  \E_n[x] \E_n[z]} = \frac{\hat{Cov}(z,y)}{\hat{Cov}(z,x)}
$$
:::info
Start here during the exam
:::

\begin{align}
\hat{\beta}_1 &= \frac{\hat{Cov}(z,y)}{\hat{Cov}(z,x)}\\
&= \frac{ \E_n[zy] - \E_n[z] \E_n[y]}{ \E_n[zx] -  \E_n[x] \E_n[z]}\\
\end{align}

Since $z$ is binary, define $\hat{p} = \frac{1}{n}\sum_{i=1}^n z_i$. By the definition,

$$
\bar{X}_1 = \frac{1}{\sum_{i=1}^n z_i} \sum_{i=1}^n z_i x_i = \frac{1}{n\hat{p}} \sum_{i=1}^n z_i x_i,\\
\bar{X}_0 = \frac{1}{\sum_{i=1}^n (1-z_i)} \sum_{i=1}^n (1-z_i) x_i = \frac{1}{n(1-\hat{p})} \sum_{i=1}^n (1-z_i) x_i
$$

Hence, 

$$
\E_n[zx]=\hat{p} \bar{X}_1, \E_n[x] = \hat{p} \bar{X}_1 + (1-\hat{p}) \bar{X}_0, \E_n[z]= \hat{p}
$$

Similarily,

$$
\E_n[zy]=\hat{p} \bar{Y}_1, \E_n[y] = \hat{p} \bar{Y}_1 + (1-\hat{p}) \bar{Y}_0, \E_n[z]= \hat{p}
$$

Finally,

$$
\hat{\beta}_1 = \frac{\hat{p} \bar{Y}_1 - (\hat{p} \bar{Y}_1 + (1-\hat{p}) \bar{Y}_0)\hat{p}}{\hat{p} \bar{X}_1 - (\hat{p} \bar{X}_1 + (1-\hat{p}) \bar{X}_0)\hat{p}} = \frac{\hat{Y}_1-\hat{Y}_0}{\bar{X}_1-\bar{X}_0}.
$$

### b
Yes.

By WLLN and Slutsky,

\begin{align}
\hat{\beta}_1 &= \frac{\hat{Cov}(z,y)}{\hat{Cov}(z,x)} \to_p \frac{Cov(z,y)}{Cov(z,x)}\\
&= \frac{Cov(z,\beta_0 + \beta_1 x +u)}{Cov(z,x)} = \frac{\beta_1 Cov(z,x)}{Cov(z,x)} = \beta_1,
\end{align}

where we use $Cov(z, u)=0$ and $Cov(z, x) > 0$ from the assumption.
# Q3

# Q4
