# Why $v_i=(X_i-\mu _X)u_i$ is i.i.d?

I don't understand. Ok, we have $\beta_1-\hat{\beta }_1=\frac{\frac{1}{n}\sum_{i=1}^{n}v_i}{(\frac{n-1}{n}){s_{X}}^{2}}$. So, for the first OLS assumption results that $E(v_i)=E((X_i-\bar{X})u_i\overset{p}{\rightarrow}E((X_i-\mu_X)u_i)=E(X_i-\mu_X)E(u_i)=0$ (since $E(u_i|X=x_i)=0$). Then Stock and Watson says that for the second OLS assumption $v_i$ is i.i.d. and $Var(v_i)<\infty$. Surely this is for the casual sampling of variables but i can't prove it clearly. After this, we have $E(v_i)=0$ and $Var(v_i)={{\sigma _{v}}^{2}}<\infty$ so we can apply CLT and saying that for $n\rightarrow \infty$ $v_i\sim N(0,{{\sigma _{v}}^{2}}/n)$ since $\frac{1}{n}\sum_{i=1}^{n}v_i=\bar{v}$.

• Please make your post more clear. What is the "second OLS assumption": for example? Also formulate your actual question more clearly. For example the i.i.d. assumption has exactly to do with the way we sample the observations, it is unrelated to the assumptions pertaining to the regressors and the error term "inside" each observation. – Alecos Papadopoulos Dec 30 '17 at 23:04
• The OLS second assumption is related to the sample extraction mode of the variables: if a sample of $n$ units is extracted from the population through simple random sampling, then the pairs of values $(X_i, Y_i)$, $\forall i$ are i.i.d. Quoting Stock and Watson: "$(X_i,Y_i), i=1,...,n$" are extracted independently and identically distributed (i.i.d.) for their joint distribution". – Francesco Totti Dec 31 '17 at 6:54
• @Alecos Papadopoulos However all this is needed to explain the asintotic distribution of $\hat{\beta_1}$: i can't explain this knowing $\beta_1-\hat{\beta_1}=\frac{\frac{1}{n}\sum_{i=1}^{n}v_i}{(\frac{n-1}{n}){s_{X}}^{2}}$. – Francesco Totti Dec 31 '17 at 12:27

Even after the comments exchange I am not totally sure what is the problem of the OP, but I will answer the question in the title.

Assume that we have a model

$$Y_i = \beta X_i + u_i,\;\;\; i=1,...,n$$

Assume that we obtain a "random sample" of observations from $\{Y_i, X_i\}$. This makes each observation $i$ identically and independently distributed from all other observations.

Now consider the random variable

$$v_i=(X_i-\mu _X)u_i = (X_i-\mu _X)(Y_i - \beta X_i) = X_iY_i - \beta X_i^2 - \mu_XY_i+\mu_x\beta X_i$$

while also

$$j\neq i : v_j=(X_j-\mu _X)u_j = (X_j-\mu _X)(Y_j - \beta X_j)=X_jY_j - \beta X_j^2 - \mu_XY_j+\mu_x\beta X_j$$

$v_i$ is a function of random variables $\{Y_i, X_i\}$ only, while $v_j$ is a function of random variables $\{Y_j, X_j\}$ only which are independent from $\{Y_i, X_i\}$. So $v_i$ is independent from $v_j$.

More over, since $X_i$ follows the same distribution as $X_j$, and $X_j$ follows the same distribution as $Y_j$, we have that $X_iY_i$ follows the same distribution as $X_jY_j$.Etc. So we also conclude that $v_i$ will have the same distribution as $v_j$.

• Thanks a lot for the correction. I understood the concept now. – Francesco Totti Dec 31 '17 at 16:21
• @FrancescoTotti. Glad to hear. Since this answer appears useful to you, please check the green mark so that your post leaves the queue of unanswered questions – Alecos Papadopoulos Dec 31 '17 at 17:04