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COVID-19

My unpublished COVID-19 paper is now my most cited paper

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Recently, my working paper Susceptible-Infected-Recovered (SIR) Dynamics of COVID-19 and Economic Impact has surpassed my JEBO paper in terms of citation counts, and has become my most cited paper. My COVID-19 paper is one of the very first written by an economist on this topic, and it appeared in the first issue of the working paper series Covid Economics. Although I am no longer working on this paper since the situation with COVID-19 has been changing too quickly (especially when I wrote the paper in March 2020) to keep up with, I am glad that this paper has made some impact. In fact, it was featured in VoxEU and Fortune articles.

Economics

My unpublished COVID-19 paper is now my most cited paper

1 minute read

Published:

Recently, my working paper Susceptible-Infected-Recovered (SIR) Dynamics of COVID-19 and Economic Impact has surpassed my JEBO paper in terms of citation counts, and has become my most cited paper. My COVID-19 paper is one of the very first written by an economist on this topic, and it appeared in the first issue of the working paper series Covid Economics. Although I am no longer working on this paper since the situation with COVID-19 has been changing too quickly (especially when I wrote the paper in March 2020) to keep up with, I am glad that this paper has made some impact. In fact, it was featured in VoxEU and Fortune articles.

LaTeX

Difference between $$..., \[..., and \begin{equation*}...

less than 1 minute read

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I have been using \(\LaTeX\) for over 20 years now. When I write a displayed equation without numbering on a single line, I have been using $$...$$ because it was simple. I didn’t understand why some people use \[...\], because the latter takes more time to type and is not necessarily easy to read. Today I read this article and learned that $$...$$ is incorrect. From now on, I will switch to \begin{equation*}...\end{equation*} because it is easy to read and we can add equation numbering by deleting * if we change our mind.

announcements

Created new website

less than 1 minute read

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I have created a new website. I have been using Google Sites to create my old website. I like the classic Google Sites because it allows the user to take control of the structure by programming in html. However, the new Google Sites no longer have this feature, and since the classic Google Sites will be discontinued in 2021, I had to do something else. After a bit of Google search, I found this template, which is exactly what I wanted (ability to take full control, free, no advertisements, etc.).

economics

food

White rice and brown rice

less than 1 minute read

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There is a restaurant called “The Bistro” on UCSD campus. Although I don’t like this restaurant because it’s basically a fusion American Asian place, sometimes I have to eat there when we take seminar speakers for lunch. Once I ordered some fried cod (neither quite fish and chips nor tempura). On the menu it said the dish comes with brown rice, so I asked the server to substitute white rice for brown rice. (Most Japanese people eat white rice - only those with strong opinions eat brown rice, though obviously the latter is healthier.) When the dish arrived, I was stunned that the rice, though white, was sushi rice (i.e., vinegared rice). I asked the server to bring proper white rice but she didn’t know the difference. Since then, whenever I organize the seminar lunch, I choose a different place.

mathematics

Inequality on spectral abscissa

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Let \(A, B\) be square (complex) matrices such that \(|B| \le A\). Then it is well known that \(\rho(B) \le \rho(|B|) \le \rho(A)\), where \(\rho\) denotes the spectral radius (largest absolute value of all eigenvalues). See, for example, Theorem 8.4.5 of Horn and Johnson (2013). In my recent paper, we needed to use the spectral abscissa (largest real part of all eigenvalues) instead of the spectral radius. By analogy, we can make the following conjecture: if \(A, B\) are square complex matrices such that \(\mathrm{Re} b_{nn} \le a_{nn}\) for all \(n\) and \(|b_{nn’}| \le a_{nn’}\) for all \(n \neq n’\), then is it true that \(\zeta(B) \le \zeta(A)\), where \(\zeta\) denotes the spectral abscissa?