581 reputation
19
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location Atlanta, GA
age 27
visits member for 1 year, 11 months
seen Aug 19 at 21:15

I am a Ph.D. student in Operations Research at the Georgia Institute of Technology. I am an avid user of C++, Mathematica, and LaTeX. I benefit a great deal from answers on the Stack Exchange network, and I hope to contribute as much as I can to the growth of this great community.


Mar
13
comment Tricks. If $\{x_n\}$ converges, then Cesaro Mean converges (S.A. pp 50 2.3.11)
I appreciate your curiosity. I will need a to think through this, and answer your revised questions soon. Research meetings this week, have me busier than usual.
Mar
10
comment Tricks. If $\{x_n\}$ converges, then Cesaro Mean converges (S.A. pp 50 2.3.11)
@TuckerRapu, You are correct that I made a typo around the word "regular". I am not 100% clear on the inequality you are referring to in question 5, so please let me know if I did not address your question. I am also not completely clear on what you mean by "We can't just choose whatever we want?".
Mar
10
revised Tricks. If $\{x_n\}$ converges, then Cesaro Mean converges (S.A. pp 50 2.3.11)
added 1422 characters in body
Mar
10
revised Tricks. If $\{x_n\}$ converges, then Cesaro Mean converges (S.A. pp 50 2.3.11)
added 1422 characters in body
Mar
5
answered Tricks. If $\{x_n\}$ converges, then Cesaro Mean converges (S.A. pp 50 2.3.11)
Sep
17
awarded  Yearling
Feb
13
answered Show that each of the following equations has a solution of the form $u(x,y) = f(ax+by) $ for a proper choice of constant $a,b$.
Feb
10
comment The Probability of an event occuring an exact amount of times
Think about how you answered question 1, and you will see that $\binom{40}{2}$ is not giving the same information. If exactly 2 people share the same birthday, that means that the remaining 38 people must have different birthdays from each other and from the two who have the same birthday. Think of a simpler case where you have three people, can you answer the question then?
Feb
10
answered The Probability of an event occuring an exact amount of times
Feb
10
comment The Probability of an event occuring an exact amount of times
If there are 40 people in the room, then there are anywhere between 1 birth day in the room (everyone has the same birthday) and 40 birth days in the room (everyone has a different birthday). Having only two people sharing the same birthday is the same as saying there are exactly 39 brithdays in the room.
Feb
6
comment Is the event $\{\max\{X_1,X_2\}=X_2\}$ measurable with respect to $\sigma(\max\{X_1,X_2\})$?
@Ilya, what I mean to say is the following. Consider my example where I use exponential random variables with $\lambda=\mu$. If $F$ is the distribution function of $Y=\max\{X_1,X_2\}$ and $G$ is the distribution function of $Z=1_{\{X_1\leq X_2\}}$, then the joint distribution function $H$ of $Y$ and $Z$ is given by the product of $F$ and $G$.
Jan
29
comment Is the event $\{\max\{X_1,X_2\}=X_2\}$ measurable with respect to $\sigma(\max\{X_1,X_2\})$?
@Ilya, I think your explanation matched what my intuition was telling me, but I had no idea how to make it rigorous. In the example that I was working, $X_1$ and $X_2$ were exponential random variables with rate $\mu$ and $\lambda$, respectively. The problem was to check the independence of $\max(X_1,X_2)$ and the indicator $1_{(X_1\leq X_2)}$. The random variables are independent if and only if $\lambda=\mu$. Someone presented a proof based on the reasoning in my question, and I couldn't verify the validity of this person's proof.
Jan
29
accepted Is the event $\{\max\{X_1,X_2\}=X_2\}$ measurable with respect to $\sigma(\max\{X_1,X_2\})$?
Jan
28
asked Is the event $\{\max\{X_1,X_2\}=X_2\}$ measurable with respect to $\sigma(\max\{X_1,X_2\})$?
Oct
3
comment Convergence in measure does not imply $L^1$ convergence
Thank you very much. I was sloppy, and too quick to answer. I mostly work in probability spaces, hence finite measure spaces, so I did not proof my work correctly. My new answer should be correct. Sorry for the confusion.
Oct
3
revised Convergence in measure does not imply $L^1$ convergence
Example was wrong
Oct
3
answered Convergence in measure does not imply $L^1$ convergence
Oct
3
awarded  Organizer
Oct
3
revised Create a unique numbers
Removed tags that do not seem appropriate.
Oct
3
suggested suggested edit on Create a unique numbers