# mike

 2d comment Blumenthal 0-1 lawyou're right, (see above comment). I had this argument in mind: look in at $n=2^j$, then you can show that there is a prob bounded away from zero of etc. (compare to geometric) , and omit 0-1 law. Dec 5 comment Blumenthal 0-1 lawI don't agree with yr first inequality, although I do agree that brownian scaling is the easy way to do this. Dec 5 awarded Supporter Dec 3 comment Does anyone know a solution to this PDE?for $\gamma = 1$ it is the generator of a squared bessel process. these were studied by fourier methods in the 50's by Feller, and are well know today as the CIR interest rate processes. Nov 27 comment A simple filtering problemsorry, i got into a typesetting mess i couldn't get out of: my suggestion is do the case $u_t=$ constant explicitly. I think they are the same, that your expression in big parens is dW under the full measure Nov 26 comment Relashionship betwen Weibull and Chi-square distributions?a weibull is a power of an exponential, which is a chi-sq Nov 21 comment Proof on Brownian Bridgecompute its covariance function Nov 20 comment Number of times above a linear boundary for a finite variance random walkit's true with 2nd moments, not fewer, and is known as hsu-robbins or hsu-robbins-erdos, and is from around 1952 Nov 19 comment "Continuity" of stochastic integral wrt Brownian motionLet me put that another way: i think the only thing that matters is that var $\int ( H_s - H_0 ) dB << \epsilon$ and its mean is 0. That 't' in previous post shouldn't have been there. Nov 19 comment "Continuity" of stochastic integral wrt Brownian motionI say apply Ito isometry to $\int ( H_s - H_0 ) dB$. Suppose using continuity you know $| H_s - H_0| < \delta$. Then the variance of $\int ( H_s - H_0 ) dB < \delta \epsilon t$ and the ratio is a (mean 0) rv with variance $\delta \epsilon t$ / a normal with variance $\epsilon t$ , and that must be small in probabiltiy. Nov 19 comment Central limit theorem and Slutskyyou can replace $B_i$ with $B_i-EB_i$, so assume it has mean 0. Then the CLT applies to the i.i.d. r.v.s $A_iB_i$, which have variance $Var(B) E(A^2)$, so you apparently you have a simple quote of slutsky to the effect of $\frac {A_1 + ... + A_n} n$ converges. Nov 19 comment "Continuity" of stochastic integral wrt Brownian motionwrite $H_s = (H_0) + (H_s - H_0) = (I) + (II)$. The contribution from I gives you what you want. Use the Ito isometry to show that the contribution from II is very small, by showing that it has a variance much smaller that $\epsilon$ Nov 17 answered Copulas, implication Nov 16 comment Difficult Discrete/Probability Problemi replace the $f_i$ with iid U(0,1) to avoid ties, and believe $E(ic(f) ) = \frac 12 {n \choose 2}$ and variance order of $n^3$. would have to get the details see how it works out, plus the messy issue of the r.v.s being uniform on 1,...,n. Nov 15 comment Difficult Discrete/Probability Problemhave you tried the expectations method ? get the mean and variance of the number of such pairs & use markovs inequality ? Nov 14 comment Exchangeability and Independence1. you don't need to find $f_i$, just observe that it is not a product. 2. for exchangeability you should show that the joint dist of (X,Y,Z) is the same as that of (Z,X,Y) is the same as ... all permutations, which is somewhat more than you are showing. Nov 14 comment $L^1$ convergence of a sequence of stochastic integrals and convergence of their quadratic variationsshould have said something along the lines of 'they are $\mathbb L^2$ bounded because their quadratic variations are $\mathbb L^1$ bounded, as they are a convergent sequence in $\mathbb L^1$. I.e., since $\mathbb E (Z_t^n)^2) = \mathbb \int_0^t (\sigma_s^n)^2 ds$ which is converging.' Nov 14 comment Autocorrelation functions of 2 correlated stationairy processesdo you know that Y is an AR(1) process with ARCH noise ? Nov 13 comment Distributions of a group of i.i.d Gaussians after Gram-Schmidt Orthogonalizationthat is the haar measure on the orthogonal group, but i don't know a better description of it than the one you just gave. Nov 12 comment $L^1$ convergence of a sequence of stochastic integrals and convergence of their quadratic variationsthe first part is certainly not right, as $\sigma_n = (-1)^n$ shows. I think the second part is correct. The martingales in question are $\mathbb L^2$ bounded by the $\mathbb L^1$, and if they converge ptws, they will also converge in $\mathbb L^p, p<2$. Does this imply $\mathbb L ^{\frac p2}$ convergence of quadratic variation by burkholder-davis-gundy ?