What does a Probability Theory PhD Qualifying Exam look like?

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Struggling Grad Student

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In question 8a), M_{t} looks like the Dolean-Dades exponential martingale, which solves the basic differential equation of geometric Brownian motion, and its expectation is always one.

sdm

Very much turning into my favorite channel, even though I can I know what some of the notation means these questions seem like they're in a different language. xD

jacoblewis

Man. I’m
So scared for this when I take this in my stats PhD

prod.kashkari

It's relieving that at least this exam seems to give some freebies (e.g. the ones about definitions), and the problems seem to be short or with hints on what theorem to apply. It's very unlike the real analysis exam which made you memorize the proofs of all different theorems and also test your problem solving abilities to the max.

FT

The sequence where X_n = n in the interval [0, 1/n) and 0 everywhere else doesn't converge to anything in L^1. The only thing it could converge to is the 0 function (after all, if there is a set of positive measure where the limit function f differs from 0 then at some point most of that set is outside the interval [0, 1/n) and the L^1 norm of X_n - f would be strictly positive for all n after that). But the function does not converge to 0 either because it would need to converge to 0 in norm and the norm of X_n is 1 all the time.

MK-

Dominican Graduate Student here from an MSc in Applied Math, taking stochastic calculus now and definitely seeing how much I have left in probability haha

manuelcastellanos

This is more in my lane, as a masters degree in mathematical finance. I had to take the PhD level probability course twice (was an optional course for me, but glad I took it), to improve my grade from D to B. About 3000 thorems to remember and be able to prove.

pandabearguy

Me just finishing calc 1 tryna understand this 😅

marvingordon

No, there's no way 5 is that easy, right?.

The X_n are given to be independent, and we are given the distributions of each which are identical. Hence let the common distribution be X. We have E(X) = 0, E(X^2) = 1/2 = Var(X).

If S_n is defined as customary (the sum of the first n random variables), then by the (classical) central limit theorem, S_n/n -> N(E(X), Var(X)/n) in distribution. Hence, S_n/sqrt(n) -> sqrt(n)N(E(X), Var(X)/n) = N(E(X)sqrt(n), Var(X)) = N(0, 1/2) in distribution

abblabaabblaba

I am finishing a bachelor degree in Statistics and in my opinion I could pass this test with 1 month of study. I thought it was way harder

lucianozaffaina