Wikipedia:Reference desk/Archives/Mathematics/2016 July 27

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July 27

Multivariate Characteristic Function

Hi, I have a somewhat peculiar question: The multivariate characteristic function is defined as $\phi _{\textbf {X}}({\textbf {t}})=E[\exp(i{\textbf {t}}\cdot {\textbf {X}})]$. Now assuming

a) We know that all multivariate Moments of ${\textbf {X}}\in \mathbb {R} ^{n}$ exist and are finite.

b) We have an explicit expression for ${\tilde {\phi }}_{\textbf {X}}({\textbf {t}})$ for ${\textbf {t}}$ in an open convex set not including zero (which immediately means we know it in an open cone $C\subset \mathbb {R} ^{n}$).

Can we conclude that the moments of ${\textbf {X}}$ are given by the derivatives of ${\tilde {\phi }}_{\textbf {X}}({\textbf {t}})$? I would claim that this is true because of the following argument, where the very first sentence is the one I am not sure about: Directional derivatives of arbitrary order satisfy $\partial _{{\textbf {v}}_{1}}\dots \partial _{{\textbf {v}}_{n}}{\tilde {\phi }}_{\textbf {X}}(0)=\partial _{{\textbf {v}}_{1}}\dots \partial _{{\textbf {v}}_{n}}\phi _{\textbf {X}}(0)$ as long as all ${\textbf {v}}_{i}$ are in $C$. As $C$ contains a basis of $\mathbb {R} ^{n}$, we therefore get the full multivariate moments of ${\textbf {X}}$ expressed in this basis in terms of derivatives of ${\tilde {\phi }}_{\textbf {X}}$. Due to the linearity of the moments these expression in a "weird" basis can simply be linearly transformed to a standard basis expression.

Is this argument flawed? It certainly does not hold simply for any function, as the function could simply not be differentiable at 0. Can we conclude from the existence of all moments that the characteristic function must be differentiable at 0? Is this sufficient to make the "linearity argument" I am making here? -- 134.76.84.240 (talk) 16:35, 27 July 2016 (UTC)

• I am not sure about the rest, but: Can we conclude from the existence of all moments that the characteristic function must be differentiable at 0?

No. For example, take $f(x)={\begin{cases}\sin(1/x)&x\neq 0\\0&x=0\end{cases}}$ on $x\in [-1,1]$ (with f(0)=0). It is bounded and has only one singularity point, so all moments are defined (integrate on $[-1,-\epsilon ]$, the result is finite, and has a finite limit as epsilon goes to zero), but the function itself is not even continuous in 0. The same result holds for nonbounded intervals, for example multiply f by any C^∞ (see Smoothness) function such that $g(x>1)=g(x<-1)=0$ yet $g(0)=1$ and then fg verifies the property on $]-\infty ,+\infty [$. (Such functions exist; a classical exercise is to prove that $h(x)={\begin{cases}e^{-1/x}&x>0\\0&x\leq 0\end{cases}}$ is infinitely derivable, you can easily craft g from that). Tigraan^{Click here to contact me} 15:45, 28 July 2016 (UTC)

Thank you for your remark, but I am afraid it might not be helpful for my problem. Your example is not a valid characteristic function because a characteristic function is always uniformly continuous. Also I am asking whether the characteristic function of a random vector ${\textbf {X}}$ is differentiable at 0, if the moments of the probability distribution of ${\textbf {X}}$ all exist. I am sorry for stating things so sloppy and not making this clear in my statement above. I could try to make a lame excuse that this sloppy jargon was commonplace among statisticians, but that would probably be an undeserved slur. -- 134.76.84.240 (talk) 14:53, 29 July 2016 (UTC)

Why is 2 · 3³⁵ so Close to 10¹⁷ ?

I am looking for an elegant approach to showing that $2\cdot 3^{35}\simeq 10^{17}.$ Approximating 3⁵ with 2⁸ or 5³ with 2⁷ doesn't really seem to do the trick, for some reason (the exponents are too large, I guess). — 79.118.178.11 (talk) 21:53, 27 July 2016 (UTC)

Try logarithms:
$\lg(2\cdot 3^{35})=\lg 2+35\cdot \lg 3\approx 17.0002739=\lg \left(10^{17.00027391}\right)$
so
$2\cdot 3^{35}\approx 10^{17}\cdot 10^{0.00027391}\approx 10^{17}\cdot 1.0006309$
You may be interested in oher such coincidences, presented in Almost integer. --CiaPan (talk) 23:19, 27 July 2016 (UTC)

For obvious reasons, I do not consider this to be an elegant approach. I'd have to add up about ten terms of the form ${\frac {1}{2^{k}k}},$ ten more of the form ${\frac {2^{k}}{3^{k}k}},$ and about fifteen of the form ${\frac {4^{k}}{5^{k}k}},$ to come up with a decent approximation. Might as well do the exponentiation by repeated squaring, computing only the first three relevant digits each step of the way, starting at 3⁴ = 81. — 79.118.178.11 (talk) 02:04, 28 July 2016 (UTC)

The values agree to about one part in 10³, and there are more than 3 digits in the inequation, so it seems likely that it's just the look-elsewhere effect, and there's no simpler demonstration of the agreement because there's no deeper reason for it. -- BenRG (talk) 03:16, 28 July 2016 (UTC)

there's no deeper reason for it — I'm not so sure about that part. After all, it all started with me searching for a number of the form 2^a · 3^b in between 2000 and 3000, then using squaring and cubing to see if by any chance I don't stumble upon something interesting, and indeed, lo and behold, the number 5281 popped up, since ${\sqrt[{9}]{5281}}\simeq 2.592={\frac {2^{2}\cdot 3^{4}}{5^{3}}}.$ Now, 5280 is a rather famous example of number of the form 2^a · 3^b · n (with n = 55), since there are, by definition, exactly 5280 feet in an English mile. Also, it lies close to 5184 = 3000₁₂, which is a number strictly of the form 2^a · 3^b (i.e., n = 1). — 79.118.178.11 (talk) 05:02, 28 July 2016 (UTC)

I agree with BenRG. There are a lot of possible combinations of (small) numbers and necessarily some integer near-equalities will pop up. I don't see how your method of discovery makes the case that there is a deeper mathematical reason for this relation. See also Mathematical coincidence. Gap9551 (talk) 05:12, 28 July 2016 (UTC)

Never mind. Apparently, it can be conveniently proven using 3⁹ ≈ 12⁴ ≈ 2 · 10⁴, where the first part of the approximation is based on the aforementioned 3⁵ ≈ 2⁸, and the second part on Newton's binomial series for (1 + 0.2)⁴ . — 79.118.178.11 (talk) 05:51, 28 July 2016 (UTC)

Resolved

Just for general interest, a couple of similar coincidences with exponents about the same size are:

• 2⋅5¹⁸≈3²⁷
• 5³³≈3²⁵⋅2³⁷

--RDBury (talk) 11:01, 28 July 2016 (UTC)

I agree completely with BenRG. If you don't find at least as many 0's or 9's in the 'coincidence' part on the right as there are digits and operations on the left hand side then really there isn't any evidence of the coincidence being anything other than random chance. Here there were three 0's on the right and six symbols on the left hand side if you include the multiply and exponentiation or four if you're really trying to be generous and forget the operations. Dmcq (talk) 15:51, 28 July 2016 (UTC)