For the i.i.d. random variable m_1,...,m_L sampled from the binomial distribution with the parameters n_1 (number of Bernoulli trials) and P (prob. of success) what is the distribution of the product \prod_i (m_i/n_1)? We can assume that n_1 is large.
One "negative" answer is coming from the fact that for large n_1 Normal distribution is a good approximation to the Binomial distribution. That normal approximation is N(mu, sigma) with mu= n_1 *P, sigma^2 = n_1 *P(1-P) as seen next
https://en.wikipedia.org/wiki/Binomial_distribution#Normal_approximation
However,
product of normal(s) is not normal according to this link
https://mathworld.wolfram.com/NormalProductDistribution.html
But, you may get the idea of what the distribution looks like for a product of 2 or 3 normal random variables ...
And, one "positive" answer may come from this link
https://stats.stackexchange.com/questions/427815/probability-mass-function-of-product-of-two-binomial-variables
Technically, a conclusion about the product m_1*m_2* ... *m_L divided by n1^L could be (according to the link) that the ratio converges in probability to P^L.
Steftcho P. Dokov For m_1 from the binomial distribution with parameters n_1 and P the distribution of the density \mu=m_1/n_1\in [0,1] is given by P(\mu)=C\exp(-n_1 D(\mu||P)) for large n_1. Here C is normalisation constant and D(Q||P) is relative entropy between probabilities Q and P. Furthermore, expanding D(\mu||P) around \mu=P we will get Gaussian but with the support on [0,1], so \prod_i (m_i/n_1) will be the product of iid rv from this "clipped" Gaussian for large n_1. Perhaps one could argue that for large n_1 the range of integration can be extended from [0,1] to (-\infty, \infty) since most of contribution comes from values of \mu close to P and we very unlikely to see -ve values of \mu (question I ask arises in combinatorial problem where -ve \mu does not make sense).
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