31,041 reputation
265118
bio website sos440.tistory.com
location Los Angeles, CA
age 26
visits member for 3 years, 3 months
seen 4 hours ago

I am currently a UCLA student, particularly interested in probability theory.

I love mathematics, especially calculating some sort of integrals and summations with inner aesthetic appealing.

Language fluency: (G:green, A:amber, R:red)

  • Korean: Native
  • English: (Reading: G, Listening : A, Speaking : R, Writing : A)
  • Japanese: (Reading : A, Listening : A, Speaking : A, Writing : R)

Jul
22
awarded  Nice Answer
Jul
16
comment Packing an infinite sequence of disks
An obvious bound is $(1 - (1/\sqrt{a}))^{2} \leq Q(a) \leq 1-(1/a)$.
Jul
15
comment product measure of weak (-*-) converging signed measure is weak-(*)-converging?
You are right, that is a very delicate part and deserves a thorough justification... I missed that point.
Jul
15
comment product measure of weak (-*-) converging signed measure is weak-(*)-converging?
Hint. Any signed measure is a difference of two finite positive measures. (Or, depending on definition, it may be written as a difference of two positive measures of which at most one is infinite.)
Jul
15
comment how I could show that: $\sum_{k=0}^n\binom{n}{k}(-1)^k\frac{n^2+n+1}{nk+n+1}\le1 $?
@Semiclassical, I am not good at counting but I guess that it may be possible to interpret $$ \sum_{k=0}^{n} \binom{n}{k}(-1)^{k} \frac{b}{ak+b} = \prod_{k=1}^{n} \left( 1 - \frac{b}{ak+b} \right) $$ using some counting argument.
Jul
15
answered how I could show that: $\sum_{k=0}^n\binom{n}{k}(-1)^k\frac{n^2+n+1}{nk+n+1}\le1 $?
Jul
13
comment variance of $W_te^{W_t}$
How about going back to basics? $W_{t} \sim \mathcal{N}(0, t)$, so you would be able to write $$\Bbb{E}[ W_{t}^{2} \exp(2W_{t})] = \int_{-\infty}^{\infty} x^{2} e^{2x} p_{t}(x) \, dx $$ where $p_{t}(x)$ is the pdf for $W_{t}$.
Jul
11
answered How to evaluate integral $\int_{0}^{\infty} \left(\frac{1-e^{-x}}{x}\right)^n dx$.
Jul
11
revised How to evaluate integral $\int_{0}^{\infty} \left(\frac{1-e^{-x}}{x}\right)^n dx$.
Simplified unnecessarily complicated software-generated LaTeX codes.
Jul
6
awarded  Nice Answer
Jul
4
comment $\frac{\partial}{\partial z}|f|^2 = ?$
Notice that $f(z)$ is always written locally as a power series. Now what can we say about $\overline{f(z)}$?
Jul
2
awarded  Curious
Jun
30
comment If $X_{i}$ are I.I.D and $n^{-1}\sum_{i=1}^{n}X_{i}$ converges a.s/in-distribution to a constant $a$ is it true that $a=\mathbb{E}[X_{i}] $?
@NewGuy, Just to make sure, have you tried Theorem 2.2.7? If you tried it and still have difficulty proving in this direction, I will show you a solution.
Jun
30
revised If $X_{i}$ are I.I.D and $n^{-1}\sum_{i=1}^{n}X_{i}$ converges a.s/in-distribution to a constant $a$ is it true that $a=\mathbb{E}[X_{i}] $?
added 98 characters in body
Jun
30
comment If $X_{i}$ are I.I.D and $n^{-1}\sum_{i=1}^{n}X_{i}$ converges a.s/in-distribution to a constant $a$ is it true that $a=\mathbb{E}[X_{i}] $?
@NewGuy, I agree that Theorem 2.3.7 is much simpler. I will modify my argument. Thank you.
Jun
30
revised If $X_{i}$ are I.I.D and $n^{-1}\sum_{i=1}^{n}X_{i}$ converges a.s/in-distribution to a constant $a$ is it true that $a=\mathbb{E}[X_{i}] $?
added 19 characters in body
Jun
30
comment If $X_{i}$ are I.I.D and $n^{-1}\sum_{i=1}^{n}X_{i}$ converges a.s/in-distribution to a constant $a$ is it true that $a=\mathbb{E}[X_{i}] $?
@ChrisJanjigian, You're right. It should be limsup instead. Thankfully, that does not affect the logic itself.
Jun
30
answered If $X_{i}$ are I.I.D and $n^{-1}\sum_{i=1}^{n}X_{i}$ converges a.s/in-distribution to a constant $a$ is it true that $a=\mathbb{E}[X_{i}] $?
Jun
29
awarded  Necromancer
Jun
28
comment Taking an integration with joint probability integrand.
Are $X$ and $Y$ assumed to be independent? Otherwise it is not generally true.