Uniqueness of Moment Generating Functions18/10/2016
Most undergraduate probability textbooks make extensive use of the result that each random variable has a unique Moment Generating Function. In particular, we can use this result to demonstrate the effect of adding or multiplying random variables. For example, the proof that the sum of two Poisson Random Variables is also a Poisson Random Variable (with mean equal to the sum of the means of the two Poissons) is much easier if we can invoke this result.
Proof for the sum of two Poisson distributions:
Suppose $X$ and $Y$ are independent Poisson Distributions with parameters $\lambda_1$ and $\lambda_2$.
We know that the $MGF$ of a Poisson Distribution with parameter $\lambda$ is $e^{ \lambda ( e^t  1)) }$.
We can now use the result that:
$MGF_{X+Y} = MGF_X * MGF_Y = e^{ \lambda_1 ( e^t  1)) } * e^{ \lambda_2 ( e^t  1)) } = e^{ (\lambda_1 +\lambda_2) ( e^t  1)) }$
Which is the $MGF$ of a Poisson Distribution with distribution $\lambda_1 + \lambda_2$ proving the result. Do we need a proof? Call me pedantic, but I never liked the fact that this uniqueness result is usually stated without proof. For me, there was always an elegance to a textbook that began with definitions and axioms, and possibly a handful of weak results which were taken as given, and then proved everything required on the way. In my experience Probability and Stats books seem particularly prone to not being selfcontained and rigorous in this way. I suspect it has something to do with the fact that probability and stats, if done 100% rigorously, are both extremely technical and difficult! It would be a shame if we had to wait until we had mastered complex analysis and measure theory before we could learn about predicting the number of black and white balls in an urn. (I'm joking there are actually interesting and useful parts to probability once you get past of those boring exercises about urns) I think my discomfort is added to by the fact that there is also something slightly disingenuous in using a very powerful result to prove a fairly trivial special case without really understanding the general result you are using. If you do this too much, you never really understand why anything is true, just that it is true. Not only is this proof not often given, I had to look pretty hard to find it anywhere.. The general proof for all random variables requires either measure theory or complex analysis and is quite involved, therefore I thought I'd just write up the result for discrete random variables. So here is a proof of the uniqueness of $MGFs$ for discrete random variables over the support $\mathbb{N}_0$. Uniqueness of MGFs Suppose $X$ and $Y$ are discrete random variables over $\{ 0 , 1, 2, ... \} $. Further suppose that: That is: Then, $$\sum_{i=0}^{\infty} e^{tx_i} f_X (i)  \sum_{j=0}^{ \infty } e^{tj} f_Y (j) = 0$$ Changing the range for the second sum: $$\sum_{i=0}^{\infty} e^{ti} f_X (i)  \sum_{i=0}^{ \infty } e^{ti} f_Y (i) = 0 $$ Bringing the two sums together: $$\sum_{i=0}^{\infty} ( e^{ti} f_X (i)  e^{ti} f_Y (i) ) = 0 $$ Rearranging: $$\sum_{i=0}^{\infty} e^{ti}(f_X (i)  f_Y (i) ) = 0 $$ We can now think of this as a power series with coefficients: $$g(i) = f_X (i)  f_Y (i)$$ i.e. $$\sum_{i=0}^{\infty} (e^t)^i g(i) = 0 $$ Which allows us to use the result that a power series which is equal to $0$ in some interval (in this case $\mathbb{R}$), must have coefficients equal to $0$ on the interval. To see this, we consider the $nth$ derivative, which allows us to recover the $nth$ coefficient.: $$f^{(n)}(0) = \sum_{k=n}^\infty \frac{k!}{(kn)!}c_k0^{kn} = n!c_{n}$$ Which gives us our result since $g(i)=0$ implies that the two functions $f_X$ and $f_Y$ are equal. Chemical Equations and Diophantine Equations17/10/2016
I was reading about SpaceX on Wikipedia the other day, just after they announced their plan to travel to Mars, and about an hour later, as can happen on Wikipedia, I found myself reading about Chemistry and feeling nostalgic about all the subjects that I used to know in detail but now have just a hazy recollection of.
Reading the material on Chemistry got me thinking that there are quite a few similarities between chemical elements and prime numbers. All chemical compounds can be reduced to a combination of chemical elements in a unique way, and all integers can be reduced to a product of prime factors in a unique way. Both serve as the building blocks from which more complicated objects are constructed. What happens if we attempt to formalise this relationship? It turns out we can actually come up with a mathematical process for balancing chemical equations. And this solution is related to the study of mathematical objects called Diophantine equations. What is a Chemical Equation? Let’s start by reviewing what a chemical is. A chemical equation is an equation which defines a chemical reaction. The left hand side has the inputs to the reaction and the right hand side has the outputs. So for example, the process of adding hydrogen and oxygen to create water can be written as: H + O > H2O This equation though is not balanced. We’ve got two hydrogens on the right, but only one on the left. Therefore there is some information missing. Balancing a Chemical Equation How do we balance a Chemical Equation? We need to define the relative amounts of the inputs and outputs so that the total numbers of each element are the same on both sides of the equation. We do this by writing an integer before each input and an integer before each output to denote the relative amounts of each molecule in the reaction. In the case of the combustion reaction above, we can see by inspection that we need two hydrogen elements and one water element to create a water molecule, that is: 2 H + O = H2O What happens though if we need to balance a more complicated Equation? For example, here is the equation for the combustion of gunpowder: KNO3 + S + C → K2S + N2 + CO2 It is not immediately obvious to me how to balance this equation. It’s not too complicated and could probably be balanced by inspection, but what if the equation was much more complicated? Chemical Elements and Prime Numbers If we take the step of associating a prime number to each element in the equation, for example, let’s make the following associations:
And we also replace the additive notation with a multiplicative notation, then we get the following mathematical equation for the combustion of gunpowder:
$(67*17*(19^3))^a * 47^b * 13^c = ((67^2)*47)^d * (17^2)^e *(13*(19^2))^f$
Equating prime factors on the right and prime factors on the left gives us a system of linear Diophantine equations, one for each prime: a = 2d a = 2e 3a = 2f b = d c = f Diophantine Equations A Diophantine equation is a polynomial equation for which we are only interested in integer solutions. Since a chemical equation must have integer coefficients, the resulting system of linear equations is actually a system of Diophantine equations. So returning to our chemical equation above, we can solve this equation using linear algebra. In this case, I put the system of linear equations into an online calculator, and we find that solutions to this system are of the following form for integers n: a = 2 n b = n c = 3 n d = n e = n f = 3n I used the following online calculator: As we are interested in integer solutions, we need to now find n such that all our coefficients take integer values. In this case, we can take n to be 1 giving us: a = 2 b = 1 c = 3 d = 1 e = 1 f = 3 Putting this back into our equation above we get: 2 KNO3 + S + 3 C → K2S + N2 + 3 CO2 Which we can see is a balanced chemical equation. This process can be extended to balance any chemical equation, no matter how complicated. 
AuthorI work as a pricing actuary at a reinsurer in London. Categories
All
Archives
April 2021
