# Compute conditional probability for Poisson binomial distribution

Consider $$X=Y_{1}+\cdots+Y_{n}$$, where $$Y_{1}, \cdots, Y_{n}$$ are $$\mathrm{n}$$ independent Bernoulli random variables with $$Y_i\sim Bernoulli (1,p_i)$$, $$i=1,2,\cdots,n$$. Then $$X$$ has a so-called Poisson-binomial distribution of parameters $$p_{1}, \cdots, p_{n}$$ and $$n$$.

Suppose $$n=20$$ and $$p_i=0.5+i/50$$, $$i=1,2,\cdots,20$$. How to compute the conditional probability of $$P(X=i |X\le 5)$$, $$i=1,2,3,4,5$$, by formula derivation or software simulation (R or Python) ?

So far, I try my best to have the following results. $$P \triangleq P(X=i |X\le 5) = \frac{P(X=i,X\le5)}{P(X\le 5)} \overset{i=1,\cdots,5}{=} \frac{P(X=i)}{P(X\le 5)}$$ Then I resort to the library(poisbinom) package in R to compute the conditional probability P

# 20 Bernouilli r.v.s
n = 20; ii = 1:n; pp = 0.5 + ii/50
library(poisbinom)
numerator = dpoisbinom(1:5, pp) #P(X=i), i=1,2,3,4,5
denominator = ppoisbinom(5, pp) #P(X<=5)
p5 = numerator/denominator
round(p5,5)
# [1] 0.00001 0.00040 0.00742 0.09509 0.89707


However, I am not sure whether my results are correct. Further, is there any other ways [formula derivation or software simulation (R or Python)] to calculate the above conditional probability?

Is there a reason you did not mention the small possibility $$X=0$$?

That being said, your rounded numbers look correct. Here is a recursion producing the same results (including $$0$$, which has a conditional probability about $$0.000000207$$)

maxi <- 20
limitn <- 5
bernprob <- function(n){ 1/2 + n/50 }
probmat <- matrix(0, nrow=maxi+1, ncol=maxi+1)    # offset to include 0
probmat[1,1] <- 1
for (i in 1:maxi){
probmat[i+1,] <-  (c(probmat[i,], 0) * (1-bernprob(i)) +
c(0, probmat[i,]) * bernprob(i)    )[-(maxi+2)]
}
condprob <- probmat[maxi+1, 1:(limitn+1)] / sum(probmat[maxi+1, 1:(limitn+1)])
names(condprob) <- 0:limitn
condprob

#            0            1            2            3            4            5
# 2.070919e-07 1.338460e-05 4.012590e-04 7.423162e-03 9.509096e-02 8.970710e-01

• As you pointed out, $X=0$ indeed has a small probability, while it just takes 1:5 into account. Very appreciated your recursion procedures and correct confirmation! @Henry Aug 21, 2022 at 13:15