The Variance Rule of $V(cX) = c^2V(X)$

Question

As a statistics major, I learned a while back that the $V(cX) = c^2 V(X)$, where $V(X)$ is the variance of $X$. However, I never quite understood why that was the case. Consider this: We have an event $X$ such that each event is independent of any previous events. Take for instance, a roll of copper wire. We know that the mean length is $150 \text{m}$ and the standard deviation is $4\text{m}$. If we want to find the variance of $5$ rolls of copper wire, we could write this as $V(5X) = 25V(X) = 400\text{m}^2$. However, an alternative way to write the problem is doing $V(X + X + X + X + X) = V(X) + \ ... \ + V(X)$ - as they are all independent. Thus, we can simplify to $5V(X)$ and get $80\text{m}^2$. This second way seems like it should be the same as the first way, but they produce different answers. Can anyone explain this? I was helping a friend out with a similar problem and the second way actually produces the correct answer, so can anyone explain why this is the case?

Answer 1

The two examples aren't the same thing. $V[5X]$ represents the variance if you scale up your random variable $X$ by a factor of $5$, while $V[X_1+X_2+X_3+X_4+X_5]$ represents the variance of $5$ independent samples of $X$.

$V[X_1+X_2+X_3+X_4+X_5] = 5V[X]$ is the variance of five independent samples of wire which each have mean $150$ meters and standard deviation $4$ meters (your example). If you create a new random variable that represents the sum of the five rolls, its mean would be $750$ and its variance $5*4^2 = 80$.

$V[5X]$ is the expected variance if your roll of copper wire were scaled up 5 times, i.e., you now produce a single roll that has a mean length of $750$ meters and standard deviation $20$ meters. That roll's length then has a variance of $400$.

Answer 2

While the other answer do the computation very well, it is helpful to have the following intuition:

$5X$ only depends on one random thing, while $S_5:=X_1+X_2+X_3+X_4+X_5$ depends on $5$ (different) independent things. Imagine these $X_i$ are coin flips ($1$ for heads and $0$ for tails). If the first coin comes up heads ($X_1=1$), it is still possible for $S_5$ to be "small" (e.g., $=1$), if the others come up tails ($X_2=\cdots=X_5=0$). However, in the $5X$ case, there is no possibility of such "cancellation": $5X$ is guaranteed to be $5$ as soon as you know that $X=1$.

So it is intuitively obvious that $5X$ should have greater variance than $S_5$.

Answer 3

If the $5$ rolls of copper wire are chosen independently of each other, then the variance of the sum of their lengths is just $5$ times the population variance.

But multiplying the random variable by $5$ means you're not looking at $5$ independently chosen rolls of wire, but rather at $5$ times the length of that one roll that appeared first. You don't have independence here: it's as if you had five identical copies of that one roll.

Answer 4

Remember that the units of the variance are $m^2.$ If you want an intuitive understanding [in addition to @BenjaminWang's Answer (+1)], it may help to compare standard deviations.

Simulation: First, multiplying single observations by $5.$ With a million iterations one can expect about two significant digits of accuracy, maybe three.

set.seed(225)
x = rnorm(10^6, 150, 4)
mean(x);  var(x);  sd(x)
[1] 149.994    # aprx E(X) = 150
[1] 16.04013   # aprx Var(X) = 16
[1] 4.005013   # aprx SD(X) = 4

Multiplying by $5:$

mean(5*x);  var(5*x);  sd(5*x)
[1] 749.9667  # aprx E(5X) = 5*150 = 750
[1] 400.3069  # aprx Var(5X) = 25Var(X) = 25(16) = 400
[1] 20.00767  # aprs SD(5X) = 5SD(X) = 5(4) = 20

hdr = "Five Times Individual Obs: NORM(750, 20)"
hist(5*x, prob=T, br=30, col="skyblue2", main=hdr)
 curve(dnorm(x, 750, 20), add=T, col="orange", lwd=2)

Second looking at $S = X_1 + X_2 + \cdots X_5:$

s = replicate(10^6, sum(rnorm(5, 150, 4)))
mean(s);  var(s);  sd(s)
[1] 750.0015  # aprx E(S) = 150 + ... + 150 = 750
[1] 80.09643  # aprx Var(S) = 16 + ... + 16 = 80
[1] 8.949661  # aprx 80^.5 = 8.944
sqrt(80)
[1] 8.944272  # exact

hdr = "Sum of 5 Indep Obs: NORM(750, 8.94)"
hist(s, prob=T, br=60, xlim=c(650,850), col="skyblue2", main=hdr)
 curve(dnorm(x, 750, sqrt(80)), add=T, col="orange", lwd=2)