17,345 reputation
1443
bio website math.ubc.ca/~israel
location Vancouver BC
age 63
visits member for 3 years, 8 months
seen 4 hours ago
Associate Professor Emeritus, University of British Columbia, and Optimization Algorithms Researcher, D-Wave Systems, Burnaby BC

1d
comment Estimating the moments of a random variable
Upper and lower bounds for $k$ might be helpful.
Nov
23
comment Probability of close approach for multivariate normal variables
Are $x$ and $y$ independent?
Nov
21
comment Norm of swapped power series in the unit disk
Almost any polynomial, suitably scaled, is either $f$ or $g$ for a counterexample.
Nov
21
answered Continuity in Banach space for non-linear maps
Nov
20
answered Efficient computation of null space of large symbolic matrices?
Nov
17
answered Does this equation has a closed-form solution for $t$? ($(1-p)\sum_{i=0}^{n}t^i = p\sum_{i=0}^{n}(1-t)^i)$)
Nov
17
comment Does this equation has a closed-form solution for $t$? ($(1-p)\sum_{i=0}^{n}t^i = p\sum_{i=0}^{n}(1-t)^i)$)
Yes, the point is that it factors, and the cubic factor is the one that contains $p$, so that's what you're actually solving.
Nov
17
comment Does this equation has a closed-form solution for $t$? ($(1-p)\sum_{i=0}^{n}t^i = p\sum_{i=0}^{n}(1-t)^i)$)
For $n=5$ you're solving the cubic $${t}^{3}-7\,p{t}^{2}+2\,{t}^{2}+7\,pt+2\,t-7\,p+1$$
Nov
13
answered How to find equilibrium of the following game?
Nov
13
comment How to find equilibrium of the following game?
I don't understand what you say about $p = 1/2$. It seems to me that $(1/2, 1/2)$ is never an equilibrium: it's better for either player to choose either $0$ or $1$ (to get payoff $1/2$) instead of payoff $1/4$ at $(1/2, 1/2)$.
Nov
12
comment A nilpotency question on $C^{*}$ algebras
I suppose you aren't interested in commutative $A$'s?
Nov
11
comment Existence of arithmetic function satisfying a certain property
I don't think you mean $n_1 + n_2 + \ldots n_k = 0$ if $n_1, \ldots, n_k \in \mathbb N$. What do you really mean?
Nov
11
comment Are there infinitely many primes p such that both p-1 and p+1 have at most 3 prime factors, counted with multiplicity?
Dickson's conjecture would imply that there are infinitely many members of the subsequence where, e.g., $(p-1)/6$ and $(p+1)/4$ are prime.
Nov
10
revised Random walks with exponential decreasing steps
added 220 characters in body
Nov
10
revised Random walks with exponential decreasing steps
added 379 characters in body
Nov
10
revised Random walks with exponential decreasing steps
added 398 characters in body
Nov
10
answered Another type of derivative, and the associated primitive
Nov
10
comment Counterexample for closed graph theorem in unmetrizable case
The Closed Graph Theorem requires some conditions on $X$ as well, e.g. that it is a barrelled space. Otherwise, for a counterexample take any closed unbounded (partially defined) operator $A$ in a Banach space $Y$, and let $X = {\mathscr D}(A)$ with the topology of $Y$.
Nov
10
comment Random walks with exponential decreasing steps
No, you won't. $1 + g - g^2 - g^3 - g^4 - g^5 = 3 - \sqrt{5} > 0$.
Nov
9
comment Random walks with exponential decreasing steps
If the walk ever fails to return when $n$ is a multiple of $3$, then it will never return.