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I was reading an example in Durrett's book: Probability : Theory and Example, 4th edition (pdf verison) (Example 3.4.7, p.112)

The scenario is as follows:

Define $Y_1,Y_2...$ be independent random variables with $P(Y_m=1) = 1/m$ and $P(Y_m=0) = 1 - 1/m$, and then we can compute the mean and variance as $$EY_m=1/m; \;\; var(Y_m)=1/m-1/m^2$$ and the author also mentioned that $var(S_n) \sim \log(n)$;

Here are my questions:

  1. I'm not very clear about the notation "$\sim$" above, what does this notation means?? and why $var(S_n) \sim \log(n)$

  2. I also wonder why $var(Y_m) = 1/m - 1/m^2$? I mean $$var(Y_m) = E(Y_m^2) - (E(Y_m))^2 = E(Y_m^2) - 1/m^2 $$ but how to make $E(Y_m^2)= 1/m?$

Thanks.

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  1. the notation is: $a_n \sim b_n \iff \frac{a_n}{b_n}\to 1$. Here, using comparison with an integral you get that $$\sum_{k=1}^n \frac1k \sim \log n$$

  2. $$\begin{align}\text{var } Y_m &= EY_m^2 - (EY_m)^2 \\&= (1/m\times 1^2 + (1-1/m)\times 0^2) - (1/m\times 1 + (1-1/m)\times 0)^2 \\&= 1/m - 1/m^2\end{align}$$

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  • $\begingroup$ Thanks for reply; @mookid ; For question 2, can I think this way: $E(Y_m^2)=1\cdot P(Y_m^2 =1) + 0 \cdot P(Y_m^2=0) = P(Y_m^2=1)$ and since $P(Y_m =1) = P(Y_m^2 = 1) = 1/m$ so we get $E(Y_m^2)=1\cdot P(Y_m^2 =1) = 1/m?$ $\endgroup$ – Fianra Oct 15 '14 at 0:16
  • $\begingroup$ this is correct! $\endgroup$ – mookid Oct 15 '14 at 0:18

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