I am calculating this example at the moment. Are my calculations right?

The random variable $Y$ has a mean of 15 and a variance of 9. $Z = 1/3 (Y − 15)$. Show that $Z$ has variance $1$.

$Z = \frac13 (Y − 15)$ $, \newcommand{\stdev}{\mathrm{stdev}}\stdev(Y)=3$

$$\stdev(Z)=\frac13 (Y − 15) = \frac13 (Y)$$ $$\stdev(Y)=3$$ $$\stdev(Z)=\frac13 Y = \frac13 3 = 1$$ $$\mathrm{var}(Z) = 1^2 = 1$$

Is my argumentation right?


1 Answer 1


You're correct, though I'm not exactly sure why you converted variances to standard deviations and then back, since you can derive your answer from two properties of variance, namely $$\newcommand{\Var}{\mathrm{Var}} \Var(X-c) = \Var(X)$$ $$ \Var(cX) = c^2 \cdot \Var(X)$$

where $X$ is a random variable and $c$ is a constant.

The first property tells us that $\Var(Y-15) = \Var(Y) = 9$. Using the second, we find that

$$ \begin{aligned} \Var\bigg(\frac{1}{3}Y\bigg) &= \bigg(\frac{1}{3}\bigg)^2 \cdot \Var(Y) \\ & = \frac{1}{9}\Var(Y) \\ & = \frac{1}{9}\cdot9 \\ & = 1 \end{aligned} $$

Combining the two is pretty trivial!

There's a nice intuition behind both of these properties if you think about variance as "spread" or "scatter". Shifting a distribution left or right (i.e., adding or subtracting a constant) has no effect on how spread apart it is. Scaling a distribution with multiplication or division, however, stretches or squishes it (respectively), which changes the spread of the distribution and thus its variance.


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