Recent content by Milky

bXn^2 ?

The variance is the same for each step that he takes, so why wouldn't the variance just still be bx^2?

Okay, so for part a, E[Xn] = 0 because with every step the mean is 0...right? For part b.. I'm still not sure. I know Var(X) = bx^2, but I don't know how to get from X to Xn. I tried using the definition of variance: E[X^2] - (E[X])^2 but it didn't get me very far

No, I copied exactly what the book said. I'm so confused!

An individual traveling on the real line is trying to reach the origin. However, the larger the desired step, the greater is the variance in the result of that step. Specifically, whenever the person is at location x, he next moves to a location having mean 0 and variance \beta x^2. Let X_n...

I(C) would be the addition of I(C_1) and I(C_2), but since one of those integrals goes in the clockwise direction, it would be I(C_1) - I(C_2) So, then the integrals are equal?

Did I do it correctly?

Essentially, I thought that's what I was doing but now I'm confused. I have revised post 5, and what I did was 1. Created two independent paths from (x,y) to (x_0,y_0) 2. Let the two paths form a closed contour C, in which C=C_2-C_1 3. Since the integral of a closed contour is zero, then C_1=C_2...

This was the notation given to me by both my book and my professor. I have to work with what he wants me to work with. Okay, I don't know where to start now. In my notes, I have that \phi(x,y)=\int_{c}(u\circ\gamma)tds would be well defined if the integral was independent of the choice...

I've done an example in complex variables earlier, which I've repasted here. If I show you my reasoning would you be able to tell me if i proved it right? Homework Statement Let D be a connected domain in R^2 and let u(x,y) be a continuous vector field defined on D. Suppose u has zero...

So essentially, if: [tex](x,y)=(x_0,y-0), \phi(x,y)=\phi(x_0,y_0)[\tex] ?

What does it mean to prove a complex function is well defined?

Oh! Duh... you get \frac{h}{\zeta-(z+h)(\zeta-z)} And that h cancels with the 1/h on the outside. Thanks! Does anyone see where the 2! in the numerator of \frac{2!}{2i\pi}}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)^3}

Homework Statement If f(z) is analytic interior to and on a simple closed contour C, then all the derivatives f^k(z) , k=1,2,3... exist in the domain D interior to C, and f^k(z)=\frac{k!}{2i\pi}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)^{k+1}} Prove for second derivative. The...

I apologize, the reason I didn't enter that into the template is because it didn't occur to me that we needed to do it using the Taylor Series until Futurebird said it - my mistake. But that is completely fine with me because now I know how to use the Cauchy Integral Formula when I'm almost sure...