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I have just learnt standard deviation and I am wondering can we replace $X$ with 5 in order to calcualte the standard deviation of $Y$.

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    $\begingroup$ Standard deviation is not a linear function. If I subtract 3 from every number, do I change how spread out the numbers are? $\endgroup$ – Glen_b -Reinstate Monica Aug 31 '16 at 10:56
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The standard deviation $\sigma$ is invariant under changes in location, and scales directly with the scale of the random variable. Thus, for constants $a$ and $b$:

$\sigma(Y)=\sigma(a \times X + b)=|a|\sigma(X)$.

In your case $a=2$ and $b=3$, then $\sigma(Y)=\sigma(2 \times X + 3)=2\sigma(X)=2 \times 5 = 10$.

https://en.wikipedia.org/wiki/Standard_deviation

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It might help to know that $\text{Var}(aX+b) = a^2\text{Var}(X)$ where $X$ is a random variable and $a$ and $b$ are constants. Standard deviation is square root of variance.

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So you have

$$Var(x) = 25$$

We know variance is additive

$$Var(y) = Var(2x-3) = Var(2x) + Var(3) = 4 \cdot Var(x) + 0 = 4 \cdot 25 = 100$$

Then

$$SD(y) = \sqrt {100} = 10$$

Which makes a lot of sense. You added a constant ($-3$) to every number in $x$: does it change the spread of the numbers? No, of course not, the elements of $x$ are equally spaced this way, just centered around another number.

You also multiplied every number $x$ by $2$: does it change the spread of the numbers? Yes, it does. How much? Exactly $2$ times, which is the result.

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The standard deviation measures the volatility of a variable. Since 3 is a constant (not a random variable) and constants have no volatility, the standard deviation of y is 10.

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    $\begingroup$ Constants can still, technically, be random variables. They're just (almost) always the same. $\endgroup$ – JDL Aug 31 '16 at 11:25

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