Independence and conditional distribution

Question

In a problem that I'm solving I find that:

"Let data (yi,xi) be sampled randomly from a two-dimensional distribution such that y|x is N(ɑ,x^2σ^2)".

Are y and x i.i.d? maybe just identically distributed but not independent?

Finally, if my model is: y = ɑ - u (where u is the error term and ɑ is a constant), and I also know that the distribution of x does not depend on either a or σ (I don't know it this assumption is needed for answer at the following question), is u independent from x?

My question is due to the fact that I thought that if the data are sampled randomly this mean that the r.v. are independent, but how is possible that the conditional distribution (his variance) depend then on x?

Answer 1

You are considering a random sample of size $N$. Each draw $i$ performed independently of any other draw $j$. Each draw is a draw of a random vector $(Y_i,X_i)$. Here is an illustration of all the stochastic vectors under consideration:

$$(Y_1,X_1),(Y_2,X_2),...,(Y_N,X_N)$$

As already mentioned $(Y_i,X_i)$ is independent of $(Y_j,X_j)$ for any pair $i \not =j$. So $(Y_1,X_1)$ is independent from $(Y_2,X_2)$. This means that $Y_2$ is independent of $Y_1$ and $X_1$ and $X_2$ is independent of $Y_1$ and $X_1$.

$$f(Y_1,X_1,Y_2,X_2) = f(Y_1,X_1)f(Y_2,X_2)$$ integrate out $X_1$ and $X_2$ to get

$$f(Y_1,Y_2) = f(Y_1)f(Y_2)$$

But for each $i$ it is NOT the case that $Y_i$ is independent of $X_i$ the could be draws from multivariate normal with non-zero covariance.

Assuming further they are all drawn from the same distribution so for all $i$ the vector $(Y_i,X_i)\sim(Y,X)$, which is simply one convenient way of saying that all the vectors $i=1,...,N$ follows same distribution.

So we have

(1) Dependence between $Y_i$ and $X_i$ for all $i$

(2) Independence for $(Y_i,X_i)$ and $(Y_j,X_j)$ for all $i\not=j$

(3) Identical distribution $(Y_i,X_i)\sim(Y,X)$ for all $i$

Define

$$Y_i = a + X_i z_i$$

with $z_i \sim \mathcal N(0,\sigma^2)$

then $Y_i \lvert X_i$ is normal and have mean $\mathbb E[Y_i \lvert X_i]$ and variance $Var(Y_i \lvert X_i)$. Calculate the mean

$$\mathbb E[Y_i \lvert X_i] = a + X_i\mathbb E[z_i \lvert X_i] = a$$ where the last step follows from assuming that $z_i$ is independent of $X_i$ so $\mathbb E[z_i \lvert X_i] = \mathbb E[z_i] = 0$.

Calculate variance

$$Var(Y_i\lvert X_i) := \mathbb E[(Y_i - \mathbb E[Y_i\lvert X_i])^2 \lvert X_i] \\ \mathbb E[(X_iz_i - \mathbb E[X_iz_i\lvert X_i])^2\lvert X_i] \\ \mathbb E[(X_i^2(z_i - \mathbb E[z_i\lvert X_i])^2\lvert X_i]$$ use independence of $z_i$ and $X_i$ to get $$= \mathbb E[(X_i^2(z_i - \mathbb E[z_i])^2\lvert X_i] \\ = X_i^2 \mathbb E[(z_i - \mathbb E[z_i])^2\lvert X_i] \\ = X_i^2 \mathbb E[(z_i - \mathbb E[z_i])^2]\\ = X_i^2 \sigma^2 $$

If your model is

$$Y_i = a + u_i$$

then $u_i = X_i z_i$ and is not independent of $X_i$.

hope that answers some of your confusion, sorry for being repetitive.