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I am reading "A Simple Model of Herd Behavior" by Banerjee (1992). A short summary of the model is the following.

There is a probability $\alpha$ that each person receives a signal telling her that the true optimal assent holding $i^{*}$ is $i^{\prime}$. The signal need not be true, and the probability that it is false is $1-\beta$. If it is false, then we assume that it is uniformly distributed on $[0,1]$ and therefore gives no information about what $i^{*}$ really is.

Futhermore, decisions are taken sequentially. If the person has no signal, then she will imitate the first decision maker and invest in the same asset If she has a signal and the first person has chosen i = 0 the she follows her signal. If she has a signal and the first person has not chosen i = 0, she has a problem: she knows that the first decision maker had a signal and this signal is as likely to be right as her own signal she is therefore indifferent between following the first decision maker’s signal and following her own signal, hence, the second person will, in this case, follow her own signal.

the decision tree is summarised in this way

Later in the paper it is stated that the expression for the probability that no one in the population chooses the right option, however large the population is given by

$$ [1-\alpha(1-\beta)]^{-1}(1-\alpha)(1-\beta) . $$

Could someone help me in the derivation of this equation? As I am somewhat stuck...Thank you in advance!

[Update]:

Could this be the solution?

The probability that $i$ is the first person who receives a signal is: $$ (1-\alpha)^{i-1} $$ The probability that $k$ people receive wrong signals in a row until someone receives no signal is: $$ \alpha^k(1-\beta)^k(1-\alpha) $$ So the probability that the first group of people who receive signals in a row all receive wrong signals is: $$ \begin{aligned} & \sum_{i=1}^{\infty}(1-\alpha)^{i-1} \sum_{k=1}^{\infty} \alpha^k(1-\beta)^k(1-\alpha) \\ & =\alpha(1-\alpha)(1-\beta) \sum_{i=0}^{\infty}(1-\alpha)^i \sum_{k=0}^{\infty} \alpha^k(1-\beta)^k \\ & =\alpha(1-\alpha)(1-\beta) \frac{1}{\alpha} \frac{1}{1-\alpha(1-\beta)} \\ & =\frac{(1-\alpha)(1-\beta)}{1-\alpha(1-\beta)} \end{aligned} $$

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    $\begingroup$ I just glanced at the paper and the only way ( as far as I can tell ) for someone to be possibly able to help would be to first read the paper from front to back slowly. So, that's probably what your goal should be: Get someone to read it. It sounds and seems really interesting at a glance so if someone else takes a look, they might read it ? Otherwise, you could email Banerjee. He might be quite thrilled that someone is reading his relatively old paper with such fervor and interest. $\endgroup$
    – mark leeds
    May 26, 2023 at 5:12
  • $\begingroup$ The Banerjee of your paper is not the same Banerjee that I was thinking it might be but it's still worth an email. I email various people all the time with questions. Some respond and some don't. I'm not so sure it's predictable. $\endgroup$
    – mark leeds
    May 26, 2023 at 5:17
  • $\begingroup$ Without having read the paper completely my first thought is that it looks similar to some sort of finite geometric series given beta and alpha are smaller than one and positive..en.wikipedia.org/wiki/Geometric_series $\endgroup$
    – T123
    May 26, 2023 at 6:25
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    $\begingroup$ Thank you all for the comments. I have uploaded a possible, intuitive, solution. Any comment on that is appreciated! $\endgroup$
    – James_
    May 26, 2023 at 10:12
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    $\begingroup$ nice job. that's gotta be correct. the argument sounds right and it's too complicated to just be a coincidence that you obtained the same expression. $\endgroup$
    – mark leeds
    May 26, 2023 at 14:28

1 Answer 1

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To get the required probability, we just need to sum the probability of all the events with the initial history of the following type. The reason being that after any initial history of this type, it is not possible to choose $i^*$ by anyone, and even those with the correct signal afterwards will ignore their signal and join the herd.:

enter image description here

where $k$ goes from $0$ to $\infty$, and $j$ can take any value from $0$ to $k$. So the required probability is: \begin{eqnarray*} && {\color{red}{\alpha(1-\beta)(1-\alpha)}}\displaystyle\sum_{k=0}^\infty\sum_{j=0}^{k}(1-\alpha)^j(\alpha(1-\beta))^{k-j} \\ &= & {\color{red}{\alpha(1-\beta)(1-\alpha)}}\displaystyle\sum_{j=0}^\infty\sum_{k=j}^{\infty}(1-\alpha)^j(\alpha(1-\beta))^{k-j}\\ &= & {\color{red}{\alpha(1-\beta)(1-\alpha)}}\displaystyle\sum_{j=0}^\infty (1-\alpha)^j \sum_{k=j}^{\infty}(\alpha(1-\beta))^{k-j} \\ &= & {\color{red}{\alpha(1-\beta)(1-\alpha)}}\displaystyle\sum_{j=0}^\infty (1-\alpha)^j \sum_{t=0}^{\infty}(\alpha(1-\beta))^{t} \\ &=& {\color{red}{\alpha(1-\beta)(1-\alpha)}}\displaystyle \left(\dfrac{1}{\alpha}\right)\left(\dfrac{1}{1-\alpha(1-\beta)}\right) \\ &=& \dfrac{(1-\beta)(1-\alpha)}{1-\alpha(1-\beta)} \end{eqnarray*}

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    $\begingroup$ Thank you very much! It's very clear! $\endgroup$
    – James_
    May 28, 2023 at 21:38
  • $\begingroup$ Most welcome :) $\endgroup$
    – Amit
    May 29, 2023 at 4:32

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