# Distribution of a binomial random variable multiplied by a constant

I'm trying to model a process in which a success is the generation of $$2$$ items. If I model the process using a binomial random variable with p equal to the probability of success, I can compute the parameters for a binomial distribution.

For example, if I generate a $$100000$$ instances of a binomial random variable with $$250000$$ trials and probability of success $$0.00065$$:

s = rbinom(100000,250000,p2)
hist(s, breaks=30,freq=FALSE)
sapply(min(s):max(s), function(q) {
points(q,dbinom(q,250000,p2),pch=19)})


This gives the the number of successes, but I want to also model the number of observations which is $$2$$ times the number of successes. I assumed I could multiple $$p$$ times $$2$$, but that clearly isn't right. That translates to twice the probability of observing a success, and not the actual number of observations.

s2 = 2*s
hist(s2, breaks=30,freq=FALSE)
sapply(min(s2):max(s2), function(q) {
+   points(q,dbinom(q,250000,2*p2),pch=19)})


The mean is correct in that it has just shifted upward by a factor of $$2$$, but the variance is wrong. Analytically, I would have expected the variance to be $$4np(1-p)$$, but that doesn't appear to be the case here.

What am I missing?

If $$Y=X$$, where $$X$$ is a binomial distribution. Notice that $$Y=2X$$ is not a binomial distribution.

In particular, $$Y$$ always take even numbers.

Its expected value is indeed $$2np$$ and its variance is $$4np(1-p)$$ but it does not follow any binomial distribution.

Remark: From the first sentence, I am not sure if $$2X$$ is what you are interested, if a success is based on two pair of events with sucess probability $$p$$ independently and there are $$n$$ such pairs, then perhaps you are interested in $$Bin(n, p^2)$$.

Edit:

You use the $$Bin(n, 2p)$$ which has variance $$n(2p)(1-2p)=2np-4np^2 < 4np-4np^2$$

• True, this is not a proper Binomial distribution. Could it not be approximated by one, however, assuming the right parameters? I would have expected my second plot to at least approximate the shape, but it appears that the variance is too small. – KirkD_CO Mar 21 at 17:23
• See the edit to show why is your variance smaller. – Siong Thye Goh Mar 21 at 17:28
• Regarding Bin(n,p^2), I think this is taken care of in the definition of a success in the original trials. p is the probability that two events have occurred simultaneously already. The probability of any single event would be sqrt(p), correct? – KirkD_CO Mar 21 at 17:31
• If the success probability of each sub event is $\sqrt{p}$. Sure. But yup, you shouldn't end up with $2p$. – Siong Thye Goh Mar 21 at 17:33
• Thank you for the explanation regarding why the variance is smaller in my plots. That is tremendously helpful! Another option is to approximate it with a normal distribution. If I use the mean = $2np$ and variance = \$4np(1-p), I get what appears to be a good approximation. The same problem holds that this is not really a normal distribution since only even values are present. – KirkD_CO Mar 21 at 17:35