I am reading the book Introduction to Probability by Joe Blitzstein. The author defines Multinomial Normal distribution as follows.

A random vector $\mathbf{X}=(X_1,X_2,\cdots, X_k)$ is said to have a Multivariate Normal distribution if every linear combination of the $X_j$ has a Normal distribution. That is, we require $t_1X_1+\cdots +t_kX_k$ to have a Normal distribution for any choice of constants $t_1,\cdots, t_k$. If $t_1X_1+\cdots +t_kX_k$ is a constant, we consider it to a Normal distribution, albeit a degenerate Normal with variance $0$.

Theorem: If $\mathbf{X}=(X_1,X_2,\cdots, X_n)$ and $\mathbf{Y}=(Y_1,Y_2,\cdots, Y_m)$ are Multivariate Normal vectors with $\mathbf{X}$ independent of $\mathbf{Y}$, then the concatenated random vector $\mathbf{W}=(X_1,\cdots,X_n,Y_1,\cdots,Y_m)$ is Multivariate Normal.

Proof: Any linear combination $s_1X_1+\cdots +s_nX_n+t_1Y_1+\cdots +t_mY_m$ is Normal since $s_1X_1+\cdots +s_nX_n$ and $t_1Y_1+\cdots +t_mY_m$ are Normal and are independent, so their sum is Normal. $\Box$

But I am having difficulty in understanding the above proof. I know that $s_1X_1+\cdots +s_nX_n$ and $t_1Y_1+\cdots +t_mY_m$ are Normal by the definition of Multivariate Normal distribution. I also know that sum of independent Normals is a Normal. But I do not understand why $s_1X_1+\cdots +s_nX_n$ and $t_1Y_1+\cdots +t_mY_m$ are independent. All I know is that since $\mathbf{X}$ and $\mathbf{Y}$ are independent, $X_i$ is independent of $Y_j$ for all $1\le i\le n$ and $1\le j \le m$.


We can use the tools in my answer to your previous question.

If $\mathbf{X}$ and $\mathbf{Y}$ are independent, then for functions $f$ and $g$, $f(\mathbf{X})$ and $g(\mathbf{Y})$ are independent. See discussion here for this statement.

Now, define $f(\mathbf{X}) = s_1X_1 + \dots + s_nX_n$ for arbitrary constants $s_1, s_2, \dots, s_n$. Similarly, define $g(\mathbf{Y}) = t_1Y_1 + \dots + t_mY_m$. Then $f(\mathbf{X})$ and $g(\mathbf{Y})$ being independent yields the result.

| cite | improve this answer | |
  • $\begingroup$ Is there any less advanced proof/ intuitive argument to show that f(X) and g(Y) are independent? This is my introductory course, and I don't know Measure Theory and Borel measures. $\endgroup$ – Supreeth Narasimhaswamy Apr 20 '18 at 12:07
  • 1
    $\begingroup$ The proof here is a tad less complicated. $\endgroup$ – Greenparker Apr 20 '18 at 13:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.