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) { 

Histogram of binomial random variable

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)})

Distribution of 2 times binomial random variable

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 $Bin(n, p^2)$.


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

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  • $\begingroup$ 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. $\endgroup$ – KirkD_CO Mar 21 at 17:23
  • $\begingroup$ See the edit to show why is your variance smaller. $\endgroup$ – Siong Thye Goh Mar 21 at 17:28
  • $\begingroup$ 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? $\endgroup$ – KirkD_CO Mar 21 at 17:31
  • $\begingroup$ If the success probability of each sub event is $\sqrt{p}$. Sure. But yup, you shouldn't end up with $2p$. $\endgroup$ – Siong Thye Goh Mar 21 at 17:33
  • $\begingroup$ 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. $\endgroup$ – KirkD_CO Mar 21 at 17:35

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