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?

  • 1
    $\begingroup$ If the distribution of $y$ explicitly depends on $x$ how can they be independent? We also certainly can't say they have the same distribution when $x$ is completely arbitrary $\endgroup$
    – dsaxton
    Commented Jan 1, 2019 at 21:49
  • $\begingroup$ @dsaxton Therefore the fact that (yi,xi) are sampled randomly does not imply independence? $\endgroup$
    – Albert
    Commented Jan 1, 2019 at 21:53

1 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:


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 $i$


$$Y_i = a + X_i z_i$$

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.

  • $\begingroup$ Oh, I have another doubt, more coherent than the previous question. Are the condition that you specified in point (2) and (3) sufficient for apply that Law of Large Number? Because in a similar exercise a read that the LLN is applicable because the data (yi,xi) is i.i.d. $\endgroup$
    – Albert
    Commented Jan 1, 2019 at 22:27
  • 1
    $\begingroup$ Pls. if this was helpfull accept the answer and upvote. The other question are better dealt with in seperate questions rather than in comments. $\endgroup$ Commented Jan 1, 2019 at 22:39
  • $\begingroup$ I am new on the forum and I need 15 reputation for upvote, as soon as I can I will do it. $\endgroup$
    – Albert
    Commented Jan 1, 2019 at 22:45
  • $\begingroup$ just accept then so question is closed. If you consider it an answer offcourse. $\endgroup$ Commented Jan 1, 2019 at 22:46

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