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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.

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1 Answer 1

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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 $Bin(1-p,n)$.

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