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 $1\le j \le m$.


1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Commented Apr 20, 2018 at 12:07
  • 1
    $\begingroup$ The proof here is a tad less complicated. $\endgroup$ Commented Apr 20, 2018 at 13:42

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