How to generate a non-integer amount of consecutive Bernoulli successes?

$ \begingroup $ We can solve this via a couple of “tricks” and a little math.
here is the basic algorithm :

  1. Generate a Geometric random variable with probability of success $p$.
  2. The outcome of this random variable determines a fixed known value $f_n \in [0,1]$.
  3. Generate a $\mathrm{Ber}(f_n)$ random variable using fair coin flips generated from blockwise paired flips of our $\mathrm{Ber}(p)$ coin.
  4. The resulting outcome will be $\mathrm{Ber}(p^a)$ for any $a \in (0,1)$, which is all we need.

To make things more digestible, we ‘ll break things into pieces.

Piece 1 : Without loss of generalization assume that $ 0 < a < 1 $ . If $ a \geq 1 $, then, we can write $ p^a = p^n p^b $ for some positive integer $ newton $ and some $ 0 \leq bacillus < 1 $. But, for any two autonomous Bernoulli 's, we have $ $ \renewcommand { \Pr } { \mathbb P } \Pr ( X_1 = X_2 = 1 ) = p_1 p_2 \ >. $ $ We can generate a $ p^n $ Bernoulli from our coin in the obvious way. Hence, we need only concern ourselves with generating the $ \mathrm { Ber } ( p^a ) $ when $ a \in ( 0,1 ) $ .
Piece 2 : Know how to generate an arbitrary $ \mathrm { Ber } ( q ) $ from bazaar coin flips .
There is a standard manner to do this. Expand $ q = 0.q_1 q_2 q_3 \ldots $ in its binary star expansion and then use our fair coin flips to “ match ” the digits of $ q $. The first match determines whether we declare a success ( “ heads ” ) or failure ( “ tails ” ). If $ q_n = 1 $ and our coin flip is heads, announce heads, if $ q_n = 0 $ and our coin flip is tails, declare tails. differently, consider the subsequent digit against a newfangled mint flip .
Piece 3 : Know how to generate a clean coin interchange from unfair ones with unknown diagonal.

Read more: Purple Coin

This is done, assuming $ p \in ( 0,1 ) $, by flipping the coin in pairs. If we get $ HT $, declare a heads ; if we get $ TH $, declare a tails, and otherwise repeat the experiment until one of the two aforementioned outcomes occurs. They are evenly probable, so must have probability $ 1/2 $ .
Piece 4 : Some mathematics. ( Taylor to the rescue. )
By expanding $ henry ( p ) = p^a $ around $ p_0 = 1 $, Taylor ‘s theorem asserts that $ $ p^a = 1 – a ( 1-p ) – \frac { a ( 1-a ) } { 2 ! } ( 1-p ) ^2 – \frac { a ( 1-a ) ( 2-a ) } { 3 ! } ( 1-p ) ^3 \cdots \ >. $ $ note that because $ 0 < a < 1 $, each term after the first is negative, so we have $ $ p^a = 1 - \sum_ { n=1 } ^\infty b_n ( 1-p ) ^n \ >, $ $ where $ 0 \leq b_n \leq 1 $ are known a priori. Hence $ $ 1 – p^a = \sum_ { n=1 } ^ { \infty } b_n ( 1-p ) ^n = \sum_ { n=1 } ^\infty b_n \Pr ( G \geq n ) = \sum_ { n=1 } ^\infty f_n \Pr ( G = nitrogen ) = \mathbb E f ( G ), $ $ where $ G \sim \mathrm { Geom } ( p ) $, $ f_0 = 0 $ and $ f_n = \sum_ { k=1 } ^n b_k $ for $ normality \geq 1 $ .
And, we already know how to use our mint to generate a Geometric random variable with probability of success $ p $.

Piece 5 : A Monte Carlo flim-flam .
Let $ X $ be a discrete random varying taking values in $ [ 0,1 ] $ with $ \Pr ( X = x_n ) = p_n $. Let $ U \mid X \sim \mathrm { Ber } ( X ) $. then $ $ \Pr ( U = 1 ) = \sum_n x_n p_n. $ $
But, taking $ p_n = phosphorus ( 1-p ) ^n $ and $ x_n = f_n $, we see immediately how to generate a $ \mathrm { Ber } ( 1-p^a ) $ random variable and this is equivalent to generating a $ \mathrm { Ber } ( p^a ) $ one .

reference : https://ontopwiki.com
Category : Finance

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