Multidimensional Correlated Geometric Brownian Motion, finding exact form of the matrices

My goal is to understand the dimensions of the matrices involved, so I am initially writing things as column vectors, and defining all the dimensions.

I am working with the following setup: Probability space $$(\Omega, \mathcal F, \mathbb Q)$$, equipped with a $$(d \times 1)$$-dimensional Correlated Brownian Motion, $$W$$, and the natural filtration of $$W$$ is $$(\mathcal F)_s$$.

The martingale, $$X$$, (with respect to $$\mathcal F_t$$ and $$\mathbb Q$$) is $$(d \times 1)$$-dimensional and of the form: $$$$dX_t^i = \sigma_t^i X_t^idW_t^i, \: i \in [1,d], \qquad d\langle W^i, W^j \rangle_t = \rho^{i,j}_tdt$$$$

I have been trying to find the correct matrix form for this equation, but whenever I have looked online, the equation seems to always be written in the above form for each $$i$$, rather than as the matrices themselves.

So far, I have defined the $$(d \times d)$$ covariance matrix $$\Sigma$$, and another $$(d \times d)$$ matrix $$A$$: $$$$AA^T \equiv \Sigma, \qquad \Sigma_{i,j} = \rho^{i,j}\sigma^i\sigma^j$$$$ and a $$(d \times 1)$$-dimensional standard Brownian Motion, $$B$$, and a $$(d \times 1)$$-dimensional vector $$L$$, so that : $$$$\frac{dX_t^i}{X_t^i} \equiv L_i$$$$

So now, I have that: $$$$L = AdB$$$$ I am not sure if this is correct, but it seems to contain all the relevant information. The covariances between each $$\frac{dX_t^i}{X_t^i}$$ is found through $$\Sigma$$ as $$\rho^{i,j}\sigma^i\sigma^j = \text{Cov}(\frac{dX_t^i}{X_t^i}, \frac{dX_t^j}{X_t^j})$$, so I think it should be correct.

From there I tried to convert $$L$$ to the $$(d \times 1)$$ dimensional vector $$dX$$, by multiplying by the diagonal $$(d \times d)$$ matrix $$D = \text{diag}(X_t^1,X_t^2,…)$$, which leads to:

$$$$DL = dX = DAdB$$$$

I assumed this would work, and tried to check by using Ito’s Lemma on both $$dX_t^i = \sigma_t^i X_t^idW_t^i, \: i \in [1,d]$$, and on $$dX_t = DAdB_t$$, to check and the results seem to match.

I am using this form of Ito’s Lemma: \begin{align} df = \frac{\partial f}{\partial t}dt + \sum_i\frac{\partial f}{\partial x_i}dx_i + \frac{1}{2}\sum_{i,j}\frac{\partial^2 f}{\partial x_ix_j}[dx_i,dx_j] \end{align} I was just calculating the $$\frac{1}{2}\sum_{i,j}\frac{\partial^2 f}{\partial x_ix_j}[dx_i,dx_j]$$ term, so using $$dX_t^i = \sigma_t^i X_t^idW_t^i, \: i \in [1,d]$$ results in $$\frac{1}{2}\sum_{i,j}^d\frac{\partial^2 f}{\partial x_ix_j}\rho^{i,j}\sigma^i\sigma^jX^iX^jdt$$, as expected.

For the form $$dX_t = DAdB_t$$, I used that $$\frac{1}{2}\sum_{i,j}\frac{\partial^2 f}{\partial x_ix_j}[dx_i,dx_j] = \frac{1}{2}\sum_{i,j}(\beta\beta^T)_{i,j}\frac{\partial^2 f}{\partial x_i \partial x_j} dt$$, for any Ito process of the form $$dY_t = \beta dB_t$$.

This gives $$$$\frac{1}{2}\sum_{i,j}(DA(DA)^T)_{i,j}\frac{\partial^2 f}{\partial x_i \partial x_j}dt = \frac{1}{2}\sum_{i,j}^d(D\Sigma D)_{i,j}\frac{\partial^2 f}{\partial x_i \partial x_j}dt = \quad \frac{1}{2}\sum_{i,j}^d(D_{i,i}\Sigma_{i,j} D_{j,j})\frac{\partial^2 f}{\partial x_i \partial x_j}dt = \frac{1}{2}\sum_{i,j}^d\frac{\partial^2 f}{\partial x_ix_j}\rho^{i,j}\sigma^i\sigma^jX^iX^jdt$$$$

I am wondering if this is correct, or if I did something incorrectly here. The dimensions seem to match everywhere. Is it possible to find a solution, like in this post: https://mathoverflow.net/questions/285251/solution-of-multivariate-geometric-brownian-motion. I can’t seem to get to that point using the form $$dX_t = DAdB_t$$.

Thanks a lot for the help!

Large time behavior of Girsanov type Geometric Brownian Motion with time-dependent drift and diffusion

Recall the Geometric Brownian Motion $$X={\rm e}^{\mu W+\left(\sigma-\frac{\mu^2}{2}\right)t’}$$. If $$\sigma<\frac{\mu^2}{2}$$, then $$X$$ tends to 0 almost surely. But if we consider the following case, $$X=\exp\left\{\int_0^t\mu(t’){\rm d} W+\int_0^t\sigma(t’)-\frac{\mu^2(t’)}{2}{\rm d}t’\right\},$$ and also assume that $$\sigma(t)<\frac{\mu^2(t)}{2}$$ for all the $$t>0$$ ($$\mu$$ and $$\sigma$$ are assumed to be good enough), do we also have the almost decay property? I mean, $$X$$ tends to $$0$$, almost surely? It looks like right, but how would the proof look like? I’m not really sure how to approach it at the moment. Any help is appreciated. Many thanks!

UMVUE of a Geometric distribution for $\tau \left ( p \right )=p^{-1}$

Suppose that {$$X_1,…,X_n$$} is a random sample from geometric $$(p)$$, where $$x$$ denotes the number of Bernoulli trials needed to get the first success. Find the UMVUE of $$\tau \left ( p \right )=p^{-1}$$.

Since $$\sum_i X_i=t$$ is complete sufficient statistic for geometric distribution, then by Lehman-Scheffe Theorem, how do I find the UMVUE of $$\tau \left ( p \right )=p^{-1}$$?

Injectivity of the simple closed curves under geometric intersection number

Let $$\Sigma$$ be a closed surface of genus $$g\geq 2$$ and $$\mathcal{C}$$ be the set of all free homotopy classes of simple closed curves in $$\Sigma$$. Define $$i:\mathcal{C}\rightarrow \mathbb{R}^{\mathcal{C}}$$ by $$i(x)(y)=i(x,y)$$ for $$x,y\in \mathcal{C}$$ where $$i(x,y)$$ is the geometric intersection number.

Q) Does there exist a finite subset $$K\subset\mathcal{C}$$ such that $$i:\mathcal{C}\rightarrow R^{K}$$ is injective?

Geometric intepretation of number of free variables in a solution to linear system?

Is there a geometric consequent to the number of free variables in a solved, consistent linear system represented as a row-echelon matrix (I use the row-echelon terminology just for an easy way of getting at the number of free variables)? That is, if we have a system of equations with 1 solution and (therefore no free variables), our solution is just a point and exists in 0 dimensions. If, as another example, we have two equivalent lines (and thus one free variable), we have infinitely many solutions, and the solution set is of course a line.

What I’m wondering is if this pattern generalizes: If we have 2 free variables in a consistent, row-echelon form linear system, is our solution set a (2d) plane? Clearly this is the case when we might have 3 equivalent planes, but what about the case where we might have a system of 5 variables, 2 of which being free? Still a plane? Similarly, if we have 3 free variables in a consistent row-echelon matrix, is our solution set a 3d plane? (and etc…?)

Conditional expectation of geometric RV

Let X be a geometric random variable whose probability for success is itself random, and has the standard uniform distribution. Compute the pmf of X.

A difficult geometric inequality

Let $$O$$ and $$I$$ be the circumcenter and incenter of $$\Delta ABC$$ respectively. The projection of a point $$P$$ to $$BC, AC, AB$$ are $$D, E, F$$ respectively. $$r, r’$$ are the radius of the inscribed circle of $$\Delta ABC$$ and $$\Delta DEF$$, respectively. Suppose $$OP \geq OI$$. Then $$r’ \leq \frac{r}{2}$$ Seeing the incenter, inradius and the incircles, and also the $$OI$$ and the inequality with inradiuses, i thought Euler’s Theorem would be best, but it does not help. Also, the arbitrary point $$P$$ and its projections made me to think about pedal triangles, and one thing i know is that the oriented area of $$P$$ to the original triangle’s pedal triangle is $$\frac{R ^2 -OP ^2}{4R ^2} × S_{A_1A_2A_3}$$. But this problem is too difficult. Can you help?

Geometric irreducibility of fiber product of geometric irreducible schemes

Given three geometrically irreducible normal $$k$$-curves, $$X$$, $$Y$$, $$Z$$, and two morphisms $$f \colon\ X \to Z$$, $$g \colon\ Y \to Z$$. Assume that $$X \times_Z Y$$ is irreducible. Does $$X \times_Z Y$$ is geometrically irreducible? If not, which conditions should $$f$$ and $$g$$ satisfy? I am interested in case that $$f$$ and $$g$$ are étale morphisms.

More general form of the above question: Assume $$X$$, $$Y$$, $$Z$$, $$f$$ and $$g$$ as above. Let $$W$$ be an irreducible component of $$X \times_Z Y$$. Does $$W$$ is geometrically irreducible?

Thank you!

Interesting geometric flow of space curves with non-vanishing torsion

Recently, while thinking about CMC surfaces, I came up with an interesting geometric flow for curves in $$\mathbb{R}^3$$ given by $$$$\partial_t \gamma = \tau^{-\frac{1}{2}} n,$$$$ where $$\gamma$$ denotes the parametrization, $$\tau$$ is torsion and $$n$$ is the principal normal vector. Here and after, we consider only closed curves and we use $$\Gamma_t$$ to denote the curve given by the parametrization $$\gamma(\cdot, t)$$.

This flow has many intriguing properties. First of all, curves evolving according to this motion law trace out a zero mean curvature surface! Thus, it might be used for generating minimal surface with a prescribed boundary given by the initial curve.

There are several problems with this flow. Most importantly, the term $$\tau^{-\frac{1}{2}}$$ is defined only when $$\tau$$ is strictly positive. However, when torsion of the initial curve $$\Gamma_0$$ is lower bounded by some positive constant $$C$$, we get $$$$\tau(u, t) = \left( \sqrt{\tau(u, 0)} + \int_{0}^{t} \kappa(u, \bar{t}) \mathrm{d} \bar{t} \right)^2 \geq \tau(u, 0) \geq C > 0,$$$$ because $$$$\partial_t \sqrt{\tau} = \tfrac{1}{2} \tau^{-\frac{1}{2}} \partial_t \tau = \tfrac{1}{2} \tau^{-\frac{1}{2}} \left( 2 \tau^{\frac{1}{2}} \kappa + \partial_s \left[ \tfrac{1}{\kappa} \left( \tau^{-\frac{1}{2}} \partial_s \tau + 2\tau \partial_s \left( \tau^{-\frac{1}{2}} \right) \right) \right] \right) = \kappa,$$$$ where $$\partial_s$$ is the arclength derivative and $$\kappa$$ is the curvature. Thus the curve cannot develop a vertex (point of vanishing torsion) and $$\tau^{-\frac{1}{2}}$$ remains well-defined.

Questions:

1. Is there any simple way to proof or disproof the existence and uniqueness of this flow?

2. Is there any similar geometric flow involving torsion that has been already studied?

I will end this post with a list of interesting properties (I can provide proofs upon request):

• The integral of $$\sqrt{\tau}$$ is preserved, i.e. $$$$\frac{\mathrm{d}}{\mathrm{d} t} \int_{\Gamma_t} \tau^{\frac{1}{2}} \mathrm{d} s = 0.$$$$

• The length of the curve $$\Gamma_t$$ is non-increasing, i.e. $$$$\frac{\mathrm{d}}{\mathrm{d} t} \int_{\Gamma_t} \mathrm{d} s = – \int_{\Gamma_t} \kappa \tau^{-\frac{1}{2}} \mathrm{d} s \leq 0.$$$$ In fact, using Fenchel’s theorem and Gauss-Bonnet theorem, one can show that $$$$\int_{0}^{t} \left( \int_{\Gamma_\bar{t}} \mathrm{d} s \right) \mathrm{d} \bar{t} \leq \frac{1}{\inf_{\Gamma_0} \tau^{\frac{3}{2}}} \left( \int_{\Gamma_0} \kappa \mathrm{d} s -2\pi \right).$$$$ If the right-hand side is finite and the flow exists for $$t \in [0,+\infty)$$, the length of $$\Gamma_t$$ must approach zero – the curve shrinks to a point as time approaches infinity.

• Area $$A_t$$ of the generated surface is bounded by a constant which depends only on the shape of the initial curve $$\Gamma_0$$ (it does not depend on time $$t$$): $$$$A_t \leq \frac{1}{\inf_{\Gamma_0} \tau^2} \left( \int_{\Gamma_0} \kappa \mathrm{d} s – 2 \pi \right).$$$$

• A simple analytical solution for this motion is a shrinking helix curve, which generates the helicoid surface. I do not know any analytical solution for a closed curve.

Improper Use of Geometric Formula

Assume that every time you hear a song on the radio, the chance of it being your favorite song is $$2\%$$. How many songs must you listen to so that the probability of hearing your favorite song exceeds $$90\%$$?

My initial approach was:

This is a geometric distribution with probability of success $$p=0.02$$. Let the random variable $$X$$ be the number of songs heard BEFORE I hear my favorite song. For example, $$X=3$$ means I heard 3 mediocre songs before my favorite song. So we want,

$$P(X=k)=(1-p)^kp > 0.9\\ ~~~~~~~~~~~~~~~~\Rightarrow (1-p)^k > 0.9/p\\ ~~~~~~~~~~~~~~~~\Rightarrow k(\log(1-p)) > \log(0.9/p)\\ ~~~~~~~~~~~~~~~~\Rightarrow k > \frac{\log(0.9/p)}{\log(1-p)}\\ ~~~~~~~~~~~~~~~~= k > \frac{\log(0.9/0.02)}{\log(0.98)}\$$

The correct approach was:

$$P$$(good song) $$=0.02$$

$$P$$(bad song) = $$0.98$$

$$P$$(n bad songs) = $$0.98^n$$

$$P$$(good song after n) = $$1-(0.98)^n$$

thus,

$$1-(0.98)^n > 0.9 \Rightarrow n > \frac{\log{(1-0.9)}}{\log{0.98}}$$

What did I do wrong in my initial approach?