So I have this one. For part a I did $P(\frac{S_n}{\sqrt n}\geq k)=P(S_n\geq \sqrt n k)\leq \frac{Var(S_n)}{nk^2}=\frac{nVar(X_1)}{nk^2}=\frac{\sigma^2}{k^2}$. Is this correct?
For part b I thought about using Borel Cantelli but Im not sure how.
#Probability exercise
cunning pine
fiery harborBOT
So I have this one. For part a I did $P(\frac{S_n}{\sqrt n}\geq k)=P(S_n\geq \sq...
lone templeBOT
NFFN
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nova topaz
So I have this one. For part a I did $P(\frac{S_n}{\sqrt n}\geq k)=P(S_n\geq \sq...
Isn't that just a raw application of the central limit theorem?
cunning pine
Isn't that just a raw application of the central limit theorem?
the central limit theorem says that $\frac{S_n}{\sigma \sqrt n}\to Y$ in distribution where $Y\sim N(0,1)$.
And convergence in distribution doesnt imply convergence in probability so Im not sure what you mean by raw application
lone templeBOT
NFFN
cunning pine
Did I do part a wrong?
nova topaz
We don't care all that much about convergence in probability, to my understanding
we only care about the convergence of the cdf
We aren't trying to show that $Z_n = \frac{S_n}{\sqrt{n}}$ verifies:
$$\lim_{n \to \infty} P(\lvert Z_n \rvert > \epsilon) = 0$$
lone templeBOT
Rion
nova topaz
We are trying to show that $F_{Z_n}(K) - P(Z_n = K)$ has a pointwise limit
lone templeBOT
Rion
nova topaz
The CLT will tell you that it is true when K is a point of continuity for the limit distribution (which should be a gaussian)
but given that the CDF of a gaussian is continuous everywhere... well... there's that
cunning pine
We are trying to show that $F_{Z_n}(K) - P(Z_n = K)$ has a pointwise limit
arent we trying to show that $1-(F_{Z_n}(K)-P(Z_n=K))$ has a limit?
lone templeBOT
NFFN
nova topaz
arent we trying to show that $1-(F_{Z_n}(K)-P(Z_n=K))$ has a limit?
yeah but that's equivalent
since we don't care all that much abou the 1-... part
right?
cunning pine
right
nova topaz
But yeah
To my understanding, convergence in distribution is actually the one you're looking for
Do double check just in case
cunning pine
so what did I do wrong in my version?
cunning pine
so if the space is complete then it would have a limit ?
nova topaz
why would it?
it's like saying: 1/n < 1
you don't exactly establish the convergence of 1/n
in the same energy: |cos(n)| < 1
cunning pine
ok I understand
nova topaz
For the second one I think borel cantelli is a sound idea
$E_n = { Z_n < K} = {Z_n \geq K}^c$
lone templeBOT
Rion
earnest flame
So I have this one. For part a I did $P(\frac{S_n}{\sqrt n}\geq k)=P(S_n\geq \sq...
part a is a sequence of numbers, that's all
it's a bounded sequence
is it monotone_?
nova topaz
The red one is sus
nova topaz
is it monotone_?
To be fair we have no idea how to establish that
given that the distribution of S_n is pretty much unknown
earnest flame
Sn is convolution of iid variables
earnest flame
didn't imply it was to be computed, rather to counter your point of distribution being unknown
sec, lemme cook
been a while since ive done measure theory stuff
nova topaz
that doesn't sound all too true
cunning pine
We are trying to show that $F_{Z_n}(K) - P(Z_n = K)$ has a pointwise limit
ok wait before we continue the central limit theorem says that $\frac{S_n}{\sigma \sqrt n}$ converges but we need $\sigma =1$
lone templeBOT
NFFN
earnest flame
that doesn't sound all too true
lemme check brb
nova topaz
well exp(-n) cos(n) is bounded and converges
but is not monotonic from a certain rank
nova topaz
**NFFN**
Doesn't matter, you can rescale after that
$Z_n = \frac{S_n}{\sigma\sqrt{n}} \sigma$
lone templeBOT
Rion
earnest flame
that doesn't sound all too true
you're right, i messed up
monotone sequence converges iff it is bounded
nova topaz
the rescaling is possible thanks to the continuous mapping theorem
earnest flame
i can do the b part but not a consequence of a
nova topaz
i can do the b part but not a consequence of a
The part b, I don't know how to do it with borel-cantelli
earnest flame
this is fatous lemma basically
nova topaz
since I can't seem to find independent events
earnest flame
fatou and clt to be precise
$$ \mathbb P\left( \limsup \frac{S_n}{\sqrt{n}} \geqslant K \right) \geqslant \limsup \mathbb P\left( \frac{S_n}{\sqrt{n}} \geqslant K\right) = \mathbb P( \mathcal N(0,\sigma^2) \geqslant K) >0 $$
lone templeBOT
aL
earnest flame
first inequality from fatou and second equality from clt