College Math Teaching

November 1, 2016

A test for the independence of random variables

Filed under: algebra, probability, statistics — Tags: , — collegemathteaching @ 10:36 pm

We are using Mathematical Statistics with Applications (7’th Ed.) by Wackerly, Mendenhall and Scheaffer for our calculus based probability and statistics course.

They present the following Theorem (5.5 in this edition)

Let Y_1 and Y_2 have a joint density f(y_1, y_2) that is positive if and only if a \leq y_1 \leq b and c \leq y_2 \leq d for constants a, b, c, d and f(y_1, y_2)=0 otherwise. Then $Y_1, Y_2 $ are independent random variables if and only if f(y_1, y_2) = g(y_1)h(y_2) where g(y_1), h(y_2) are non-negative functions of y_1, y_2 alone (respectively).

Ok, that is fine as it goes, but then they apply the above theorem to the joint density function: f(y_1, y_2) = 2y_1 for (y_1,y_2) \in [0,1] \times [0,1] and 0 otherwise. Do you see the problem? Technically speaking, the theorem doesn’t apply as f(y_1, y_2) is NOT positive if and only if (y_1, y_2) is in some closed rectangle.

It isn’t that hard to fix, I don’t think.

Now there is the density function f(y_1, y_2) = y_1 + y_2 on [0,1] \times [0,1] and zero elsewhere. Here, Y_1, Y_2 are not independent.

But how does one KNOW that y_1 + y_2 \neq g(y_1)h(y_2) ?

I played around a bit and came up with the following:

Statement: \sum^{n}_{i=1} a_i(x_i)^{r_i} \neq f_1(x_1)f_2(x_2).....f_n(x_n) (note: assume r_i \in \{1,2,3,....\}, a_i \neq 0

Proof of the statement: substitute x_2 =x_3 = x_4....=x_n = 0 into both sides to obtain a_1 x_1^{r_1} = f_1(x_1)(f_2(0)f_3(0)...f_n(0)) Now none of the f_k(0) = 0 else function equality would be impossible. The same argument shows that a_2 x_2^{r_2} = f_2(x_2)f_1(0)f_3(0)f_4(0)...f_n(0) with none of the f_k(0) = 0.

Now substitute x_1=x_2 =x_3 = x_4....=x_n = 0 into both sides and get 0 = f_1(0)f_2(0)f_3(0)f_4(0)...f_n(0) but no factor on the right hand side can be zero.

This is hardly profound but I admit that I’ve been negligent in pointing this out to classes.


Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Create a free website or blog at

%d bloggers like this: