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I have a Binomial(n,p) random variable that is multiplied by a constant, specifically 1/n. I know for some random variables multiplying by a constant doesn't change the distribution, i.e. the Normal, but for others it does, i.e. the Poisson. I can't find anything about the Binomial when it is multiplied by a constant. I know the expected value and variance but what I'm really interested in is whether this is a valid (i.e. not improper) random variable for all constants or just for some. I would appreciate any guidance.

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It's a valid random variate for all constants, even zero (for which the probability of observing zero equals 1.) It's just a Binomial random variate multiplied by a constant; that doesn't change the fact that the probabilities are all nonnegative and sum to one.

Let us denote the constant by $c$ and the random variate itself by $x$. The original, binomial, random variate is multiplied by $c$ to get $x$. The probability distribution of $x$ is:

$$p(x;n,p,c) ={n \choose x/c}p^{x/c}(1-p)^{n-x/c}$$

except for $c=0$ of course. This works because $x/c$ transforms $x$ back to the original Binomial variate, which, in the case of discrete random variables, is all you need.

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  • $\begingroup$ Thanks for your answer. How did you derive the pmf, it would be great to see how that was done? Also, one of the values that got me thinking about this was c=1/n. For this case would the support only be x=1, because if c=1/n I would have n choose nx and I can only choose n objects out of n objects. I'm just having a hard time understanding this for constants that are fractions. $\endgroup$ – user211687 Jun 15 '18 at 15:28
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    $\begingroup$ The support would be $\{0, 1/n, 2/n, \dots, 1\}$ if $c = 1/n$. You don't have to have an integer value for a discrete random variable, just a countable number of possible values; whether they are integer values or not is irrelevant. $\endgroup$ – jbowman Jun 15 '18 at 19:01

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