# Help interpret exercise 4.2 in Probability Essentials, Jacod & Protter

Some years ago there was a question here regarding this exercise. I find it a bit confusing and I'm struggling to understand what is in fact being asked here.

My issue is with the first part of question 4.2.

I want to prove $$q_k$$ are the probabilities of singletons for a $$Binomial(1-p,n)$$.

In order to check if what is being asked can be true I computed the singleton probabilities $$p_k$$ for a $$Binomial(0.3, 4)$$ in Julia and subtracted each $$p_k$$ to $$1$$ to get the corresponding $$q_k$$. I then computed the singleton probabilities of a $$Binomial(1-0.3, 4)$$. As you can see results are not the same:

using Distributions

n = 4
p = 0.3

d1 = Binomial(n, p)
p_ks = pdf(d1)  # returns the probabilities of singletons for a B(p=0.3, n=4)
q_ks = 1 .- p_ks

d2 = Binomial(n, 1-p)
d2_p_ks = pdf(d2) # returns the probabilities of singletons for a B(p=0.7, n=4)

julia> q_ks
5-element Array{Float64,1}:
0.7599
0.5884
0.7353999999999998
0.9244
0.9919

julia> d2_p_ks
5-element Array{Float64,1}:
0.008100000000000003
0.07560000000000003
0.2646000000000001
0.4115999999999999
0.24009999999999992


Is the exercise wrong or am I missing something? Summing every $$q_k$$ doesn't even add up to 1.

Of course there is something going wrong.

Suppose that $$p_k$$ is the probability of exactly $$k$$ successes, which is what $$P(\{k\})$$ seems to indicate, and what the other answers talk about. Then, in particular we have $$\sum_{k=0}^n p_k = 1$$, because in a binomial experiment we can have exactly $$0,1,2,...,n$$ successes with probability $$p_0,p_1,...,p_n$$ respectively, and these events are mutually disjoint and exhaust the entire sample space.

However, if this is true, then $$\sum_{k=1}^n q_k = \sum_{k=1}^n (1-p_k) = \sum_{k=0}^n 1 - \sum_{k=0}^n p_k = n$$. So how can $$q_k$$ be probabilities, if their sum isn't $$1$$ always? (Note , for example that the entries in the first array $$1-p_k$$ add up to $$4$$.)

In that case there is an issue with the exercise. Here's how to rectify it.

You have written two arrays down. The first is $$1 - P(\{k\})$$ where $$P$$ is the probability assignment for $$Bin(p,n)$$. The second array is $$P'(\{k\})$$ where $$P'$$ is the probability assignment for $$Bin(1-p,n)$$.

I want you to observe that for each $$k$$, we have $$P'(\{4-k\}) + (1-P(\{k\})) = 1$$ for each $$k$$. In words, if you add up the $$k$$th first entry and $$k$$th last entry of the first and second array respectively, you always get $$1$$. For example, $$0.7599..+0.2400.. = 1$$ (first entry and last entry of first and second array resp.) and $$0.9244... + 0.756... = 1$$ (fourth and fourth last entry of first and second array resp.).

From this we get $$P'(\{4-k\}) = P(\{k\})$$. In words : if I change $$p$$ to $$1-p$$, then the probability of $$k$$ successes in the first case, equals the probability of $$n-k$$ successes in the second case, for any $$0 \leq k \leq n$$.

To correct the exercise :

If $$p_k = P(\{k\})$$ are the success probabilities for $$Bin(p,n)$$ and $$\boxed{q_k = p_{n-k}}$$, then $$q_k$$ are the success probabilities for the random variable $$Bin(1-p,n)$$.

To answer this question, the exercise is wrong, and the first array is not to be computed at all, rather a different expression was required.

EDIT : The corrected exercise is almost a formality, and follows from the equivalence of $$\binom nk = \binom n{n-k}$$ for any $$0 \leq k \leq n$$. The Poisson approximation follows by noticing similar complement relations between $$Bin(p,n)$$ and $$Bin(1-p,n)$$.