# Variance of exponential distribution when changing units

I was tutoring a Stat student taking a probability course and I came across a problem dealing with variance of exponential that has left me confused.

$X \sim \mathrm{Exp}(\mathrm{rate}=\lambda)$ so $E(X)=1/\lambda$ and $\newcommand{\Var}{\mathrm{Var}}\Var(X)=1/\lambda^2$

Customers enter a store at a rate of 2 per hour, a customer has just arrived, what is the variability of the amount of time until the next customer arrives in minutes?

$\lambda=2, \; E(X)=1/2, \; \Var(X)=1/2^2=1/4$

Here the expected value = 30 minutes (1/2 of an hour) and the variance = 15 minutes (1/4 of an hour)

Now work on minutes scale:

$\lambda=2/60=1/30, \; E(X)=30, \; \Var(X)=30^2=900$

Here the expected value = 30 minutes but the variance = 900 minutes

I am confused why I am not getting the same variance when I switch from minutes to hour and vice versa.

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The units of variance are squared: one squared hour = $60^2=3600$ squared minutes. It's exactly like converting from square miles to square feet. –  whuber Jul 13 '12 at 20:07
Ah yes. So we have 1/4 hours squared and 900 minutes squared. These should be equal, can someone show the derivation? –  Glen Jul 13 '12 at 20:26
(1/4)*3600 = 900. –  jbowman Jul 13 '12 at 20:35
And then 900/3600=1/4, thanks, too simple, I guess I'm over thinking things. –  Glen Jul 13 '12 at 22:21
Units are a reason people often use the standard deviation instead of the variance (even though it is easy to convert back and forth). The standard deviation has the same units as the random variable. –  Douglas Zare Jul 14 '12 at 9:18
Another way of looking at this is that if $X$ is the exponential variable in units of hours then $Y =cX$ converts $X$ to some other unit. $\newcommand{\Var}{\mathrm{Var}}\Var(Y) = c^2 \Var(X)$.
In your case $\Var(X)=1/4$ and $c=60$. So $\Var(Y)=3600$ and $\Var(X)=3600/4=900$.