The question generate event with $6.75\space p^2q$, $20\space p^3q^2$, $3.9\space pq$? prompted me to think more generally about which expressions of this form can be a probability in a finite coin experiment. Since the answer turns out to be rather nice, I thought I'd post it here as an answer to the general question.

We have a coin that shows heads with probability $p$ and tails with probability $q:=1-p$, and the question is whether for given $\mu\in\mathbb R^+$ and $i,j\in\mathbb N$ (not both zero) there is an event in an experiment with a fixed finite number of tosses of the coin whose probability is $\mu p^iq^j$ for all $p\in[0,1]$.


The monomial $\mu p^iq^j$ is a probability in a finite coin experiment if and only if $\mu$ is an integer and $\mu p^iq^j\lt1$ for all $p\in[0,1]$.

First, $\mu$ needs to be an integer: The probability of any event in an experiment with a fixed finite number of tosses is a polynomial in $p$ with integer coefficients; and $\mu p^iq^j=\mu p^i(1-p)^j$ is a polynomial in $p$ with coefficient $\mu$ for $p^i$, so $\mu$ must be integer for the two to be equal.

The rest of the “only if” direction is also readily proved. Since $\mu p^iq^j$ is a probability for all $p$, it must be $\le1$ for all $p$. If there were $p$ such that $\mu p^iq^j=1$, the event would have to include all elementary events, which implies $\mu p^iq^j=1$ for all $p$, which only occurs in the trivial case $i=j=0$ that was excluded in the question.

Now, to prove the “if” direction, consider an experiment with $i+j+l$ tosses. If we can select $\mu\binom lk$ elementary events with $i+k$ heads and $j+l-k$ tails for all $k=0,\ldots,l$, then the probability of the event composed of all these elementary events is

$$ \sum_{k=0}^l\mu\binom lkp^{i+k}q^{j+l-k}=\mu p^iq^j(p+q)^l=\mu p^iq^j\;. $$

There are $\binom{i+j+l}{i+k}$ such elementary events, so we can choose $\mu\binom lk$ of them if

$$ \mu\binom lk\le\binom{i+j+l}{i+k} $$

for all $k$. If we cancel factors in the three pairs of corresponding factorials, this becomes

$$ \mu(k+i)\cdots(k+1)(l-k+j)\cdots(l-k+1)\le(l+i+j)\cdots(l+1)\;, $$

and then dividing through by $l^{i+j}$ and denoting $\frac kl\in[0,1]$ by $\alpha$ leads to

$$ \mu\left(\alpha+\frac il\right)\cdots\left(\alpha+\frac1l\right)\left(1-\alpha+\frac jl\right)\cdots\left(1-\alpha+\frac 1l\right)\le\left(1+\frac{i+j}l\right)\cdots\left(1+\frac1l\right) $$

and thus to

$$ \mu\alpha^i(1-\alpha)^j+O\left(\frac{(i+j)^2}l\right)\le1\;. $$

Thus, we can satisfy this inequality for sufficiently large $l$ if $\mu\alpha^i(1-\alpha)^j\lt1$ for all $\alpha\in[0,1]$.

  • $\begingroup$ but, if you take an event from a subset of the (finite) probability space, can't $\mu$ be rational ? and even greater than one , keeping safe that $\mu p^i q^j <1$ ? $\endgroup$ – G Cab Jul 16 '18 at 21:49
  • $\begingroup$ @GCab: $\mu$ can certainly be greater than $1$; in the original question that I linked to, it was $20$. But I don't understand your question about $\mu$ being rational. Is there an error in my proof that it can't? Perhaps I don't understand what you mean by "taking an event from a subset". $\endgroup$ – joriki Jul 16 '18 at 21:54
  • $\begingroup$ My doubt was about introducing order. But I got your point now, in a different way: let's say that each of all the possible binary strings of length $n$ has a probability that corresponds to the product of $(p_1p_2\cdots p_m,\;q_1q_2\cdots q_{n-m})$ in whichever order. So $p_iq_j$ is the .."granularity" of any event you can define in them, and so $\mu$ must be an integer. (+1) $\endgroup$ – G Cab Jul 16 '18 at 23:50
  • $\begingroup$ @GCab: I don't think the argument works in quite that way. As the proof here and already the example given in the linked question illustrate, you can make use of $p+q=1$ to combine monomials into lower ones; and you can also use it to artificially increase the degree of a polynomial in $p$ and $q$; so I don't see how working with $p$ and $q$ allows for such a "granularity" argument; I think you really do need to switch to a representation in terms of $p$ alone, as I did in the proof, in order to make that argument go through. $\endgroup$ – joriki Jul 17 '18 at 3:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.