TT's Maths Thread

[TrueTears]

Hmm since it's a book for engineers I can't comment much on it basically I just flip through it and skim sections here and there, although I had a look at its introductory ODE section, pretty good and solid however definitely comes from an engineers perspective lol. Other than that, the sections on laplace and fourier are good enough for me, haven't really had a look at the other sections.

[TrueTears]

Cbf typing it again on AN as the latex codes are different, if the picture is too small, original sauce: http://math.stackexchange.com/questions/319254/conversion-of-primal-problem-into-dual-optimization-problem#319254

[Dedicated]

Glancing over the posts and there is some awesome maths in here despite not understanding it. Hope I can reach this level soon through lots of reading.

[TrueTears]

question attached

[#1procrastinator]

What area of math is this? OO

[TrueTears]

If X and Y are independent random variables (on a well defined probability space), then are f(X) and g(Y) also independent for any given real functions f and g? If so, any ideas how to go about proving it (non-measure theory methods?)

[kamil9876]

Hrmm, what if are both the zero function? Then both and are the same random variable.

[TrueTears]

Thanks

[kamil9876]

No, you can't take . For example let's look at 1 by 1 matrices i.e real numbers

but does ?

The relationship between P and C is as follows:



So in fact we can take .
 

[TrueTears]

thanks kamil

[TrueTears]

Let be a n by k matrix with full column rank, that is, rank k, hence has full rank k and is invertible. Let be a m by k matrix with full row rank, that is, rank m. Show that is of rank m.

EDIT: Actually just show that is a positive definite matrix.

kamil?

[kamil9876]

Ok so showing that it is positive definite will already imply it has rull rank .

Now we know that is positive definite. Thus we can by the theorems in your posts above find an invertible square matrix such that . Now:

where .

So it only remains to check that is of rank but this follows because C is an invertible square matrix and R is of rank m.


Further Question:
It's interesting to see if the statement about the rank holds in other fields (where of course the notion of positive definite doesn't necessarily exist).

Edit: Actually it doesn't in the field with two elements . Take , then  

[TrueTears]

ahhh thanks kamil, i just managed to prove it as well, although your way is much easier, i just used the definition of positive definiteness:

let be a m by 1 matrix. So we need to show , since is PD, then we can decompose , thus we have , now let and we have

[TrueTears]

How do we define independence for random vectors?

Eg, let be a random n by 1 vector and be a random k by 1 vector (where n does not necessarily have to equal to k), then how do we define independence between and ?

For example, if we take just 2 random variables, X, Y then X and Y are independent iff f(x,y) = f(x)f(y) where f(x,y) characterizes their joint pdf and f(x), f(y) are their marginal pdfs. If we adapt this onto the random vector case, we have , examining the RHS, we see that is simply the joint distribution of and likewise, is the joint distribution of , but then what is ? How is it defined?

Continuing on, for a more special case, consider when is jointly multivariate normal, ie, and is also jointly multivariate normal, ie, . Now in the special case of when two jointly normally distributed variables X and Y, a sufficient condition for independence is when cov(X,Y) = 0 where cov(.) denotes the covariance between X and Y. How then, do we generalise that into the random vector case? Ie, can we say that and are independent iff ? But then we would have to show and are jointly normally distributed... but this doesn't really make sense since and are already itself jointly normally distributed, so then wouldn't we be talking about the 'joint' distribution of two joint distributions?

EDIT: Thanks Ahmad for clarifying on IRC

[kamil9876]

^ In the middle of a Saturday night