4 out of 5 dentists recommend this WordPress.com site

## Month: May, 2019

### Effective Altruism- May

I turned *way too old* earlier this month. Hence, on my birthday month, I would like to record the donation I made to Effective Altruism:

I have also started reading on poverty in India. The first paper that I perused (very) partially is this.

It is a paper written by two Indian PhD students at Columbia University, who talk about the fact that there are basically two poverty lines used by the government of India. The latest one is in fact a harsher scale of poverty, and according to both such lines poverty in India has been steadily decreasing, especially in the 2004-2005 and the 2009-2010 period. The authors do not use the 5-year studies on poverty as a basis for their conclusions, but the annual expenditure survey done by the government of India. The basis for their choice is the following: people who spend more are probably earning more, and vice-versa. Hence, whether people are above or below the poverty line can be easily approximated by how much they’re spending.

I didn’t complete reading the paper for the following reasons: it seems motivated at the very outset to show that India is “shining”, it is more an instance of statistical jugglery than a commentary on the causes of poverty, and bases its conclusions on poverty lines that I don’t take to be credible. Every government is motivated to suppress data on poverty, or introduce measures of poverty that suggests that there are less poor people in their country than there really are. And I find the two poverty lines to not be a good measure for poverty in India.

I then found this book written up by people at the World Bank.

I will try to peruse relevant sections of this book and complete this article by (hopefully) this weekend

### Putnam 1994, Question 1

Just want to record a solution to a relatively simple Putnam problem here. The problem is the following:

Show that a sequence $a_1,a_2,a_2,\dots$ satisfies the condition $0. Prove that $\sum a_i$ diverges.

I tried attacking it with the usual methods of showing that a sequence diverges, but nothing seemed to work. However, every good Olympiad worth its salt asks for some experimentation, which will then yield important observations/patterns. I did the same here.

Without loss of generality, let $a_1=1$. Then $a_2+a_3\geq a_1=1$. Hence, $a_2+a_3\geq 1$. Similary, $a_4+a_5\geq a_2$ and $a_6+a_7\geq a_3$. Therefore, $a_4+a_5+a_6+a_7\geq a_2+a_3\geq 1$. Continuing this pattern, we observe that for all $n\in\Bbb{N}$, $a_{2^n}+\dots+a_{2^{n+1}-1}\geq 1$. Hence $\sum a_i$, which can be thought of as $\sum\limits_{n=1}^{\infty}(a_{2^n}+\dots+a_{2^{n+1}-1})$, diverges.