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location Houston, TX
age 32
visits member for 2 years, 5 months
seen Oct 4 at 0:35

BBA Finance/Economics, MSc Statistics, and currently pursuing PhD in Finance


Sep
30
awarded  Explainer
Jul
2
awarded  Curious
Jun
22
awarded  Popular Question
May
19
awarded  Yearling
Feb
5
awarded  Popular Question
Oct
13
accepted Exponentials of stochastic processes and Brownian motions
Oct
13
comment Exponentials of stochastic processes and Brownian motions
so the first two blanks are just $e^v rdt$ and $rdt$?
Oct
13
comment Exponentials of stochastic processes and Brownian motions
I added some more thoughts/work above.
Oct
13
revised Exponentials of stochastic processes and Brownian motions
added 362 characters in body
Oct
13
comment Exponentials of stochastic processes and Brownian motions
Also, for part 2, I will add some thoughts:
Oct
13
comment Exponentials of stochastic processes and Brownian motions
I guess not well enough.For the first question, I have a result that $dR/R=rdt$ but I don't understand the derivation.
Oct
13
asked Exponentials of stochastic processes and Brownian motions
Sep
13
awarded  Popular Question
Sep
10
accepted Proof of $E(X)=a$ when $a$ is a point of symmetry
Sep
10
comment Proof of $E(X)=a$ when $a$ is a point of symmetry
Thank you again. The second approach is not one which I have seen before, but it is simple and elegant.
Sep
10
comment Proof of $E(X)=a$ when $a$ is a point of symmetry
Excellent and clear answer. Thank you for that. The only place I am not clear is why the second to last line is z=$\infty$. I notice that you begin by assuming that the expectation exists. Is it even possible to show/prove the existence in general? I assume not since there are obvious examples of symmetric pdfs both with and without finite expectation
Sep
10
comment Proof of $E(X)=a$ when $a$ is a point of symmetry
@AndréNicolas, I agree but is there a way to prove that $E(X)$ exists in general, or must we know the form of the pdf or cdf to proceed?
Sep
10
asked Proof of $E(X)=a$ when $a$ is a point of symmetry
Sep
8
answered If $F_X(z)>F_Y(z)$ for all $z$, then $P[X<Y]>0$.
Sep
7
answered Required sample size to reduce the interval size by half