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In the paper I am reading, I come across $$ q(s) \propto \left( \frac{b}{c} \right)^{s}\quad s=\{0,1\}, \quad(1) $$ and the author says this is a Bernoulli distribution. ($b>0$ and $c>0$)

I know Bernoulli distribution is $$ p^k (1-p)^{1-k} \quad k=\{0,1\}. $$

How can we understand Eq.(1) as a Bernoulli distribution?

user2978524
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    A Bernoulli random variable is a random variable taking values $0$ and $1$. What is $s$? Is it $a$? – Stéphane Laurent Jun 21 '18 at 14:05
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    A Bernoulli distribution is one that assigns one probability $p$ to $1$ and the complementary probability $1-p$ to $0$. Letting the constant of proportionality be $1-p,$ it is immediate that the probabilities for $0,1$ are proportional to $1$ and to some number (equal to the odds $p/(1-p)$), respectively. Even though we aren't told what $a,$ $b,$ or $c$ represent, isn't it clear that formula $(1)$ defines precisely two such probabilities for the two values of $s$? – whuber Jun 21 '18 at 15:35

2 Answers2

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If you divide your formula for the Bernouilli distribution,$f(k) = p^k(1-p)^{1-k}$, by the constant $f(0)$ then you get $$f(k) \propto \frac{f(k)}{f(0)} = \frac{p^k(1-p)^{1-k}}{(1-p)} =p^k(1-p)^{-k} = \left(\frac{p}{1-p}\right)^k$$

Which works with your expression for $q(s)$ when $b=dp$ and $c=d(1-p)$ for any $d$.

Sextus Empiricus
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For the Bernoulli distribution, as you defined, you have that,

$$ P(K = 0) = p^0 (1 - p)^{1 - 0} = 1 \times (1 - p) = (1 - p) $$ $$ P(K = 1) = p^1 (1 - p)^{1 - 1} = 1 \times p = p $$

And $0 < p < 1$, then $ 0 < 1 - p < 1$. So, you can call $p = b/c$. Is this case, when $s = 0$, $q(a) \propto 1$, meaning that $(1-p) \propto 1$, which it really is. Also, we don't know if $b > c$, we also don't know if $b/c > 1$ or $b/c < 1$, them we have another function for the proportional symbol $\propto$.

I don't think this is the exact answer but it is close.

Bruna w
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