# Variance when changing from one random discrete variable to another

There 16 people divided to 8 pairs, each pair does the same task and has $$p=0.8$$ chance of completing it successfully, independent of the results of the other pairs

1. find Var(X), where X is the number of pairs which completed the task successfully
2. find Var(Y), where Y is the number of people which completed the task successfully

For (1) this is a classic binomial random variable so $$Var(X)= np(1-p) = 8 * (0.8)(1-0.8) = 0.128$$

As for (2), instinctively, I want to say that $$Var(Y) = 2 * Var(X)$$ because the $$n=16$$ instead of $$8$$ and the probability that a person completes the task remains $$0.8$$. But on the other hand I thought that I can look at $$Y$$ as twice the number of pairs that completed the task so $$Var(Y)=Var(2\cdot X)=4\cdot Var(X) = 4 * 1.28 = 5.12$$ but I'm not sure which way is correct

• The second way is correct. Commented Nov 28, 2020 at 21:02
• Definition of success for Pair is ambiguous. Commented Nov 29, 2020 at 0:18

To illustrate, I will describe a scenario consistent with my interpretation of your question. Suppose the task is for each person to roll a fair ten-sided die numbered from $$1$$ to $$10$$ inclusive, and the task is successful if the sum of the numbers they roll is not divisible by $$5$$. For example, if the numbers rolled are $$4$$ and $$2$$, then they succeeded, but if they roll $$10$$ and $$5$$ they failed. You can check that the probability this occurs is exactly $$p = 0.8$$. Then they can only both succeed or both fail.
In this case, the number of people succeeding can only be among the set $$Y \in \{0, 2, 4, 6, \ldots, 16\}$$, whereas the number of pairs succeeding can be among the set $$X \in \{0, 1, 2, \ldots, 8\}$$. Moreover, $$Y = 2X$$ for all outcomes of $$X$$.
It is not true that $$\operatorname{Var}[Y] = 2 \operatorname{Var}[X]$$. This is wrong. For a scalar constant $$c$$, we have $$\operatorname{Var}[cX] = c^2 \operatorname{Var}[X],$$ and for $$c = 2$$, this means the variance of $$Y$$ is four times the variance of $$X$$. The flaw in your first line of reasoning is that $$Y$$ is not itself binomial on $$\{0, 1, 2, \ldots, 16\}$$, because it is not possible to observe an odd number of people succeeding. Therefore, $$\operatorname{Var}[Y] \ne 16 p(1-p)$$.