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Let X be a continuous random variable with the density function:

$f_x(x) = \begin{cases} x+1, & \text{if}\ -1\leq x \leq 0 \\ -x +1, & \text{if}\ 0 \leq x < 1\\ 0 & \text{otherwise} \end{cases} $

The expected value of this is 0, as $E[x] = \int_{-1}^{0}x(x+1)dx + \int_{0}^{1}x(-x+1)dx = 0$

My question is: When calculating the variance using: $Var(X) = E[X^2] - (E[X])^2$

Do I simply compute the above-mentioned but with this distribution function:

$f_x(x^2) = \begin{cases} x^2+1, & \text{if}\ -1\leq x \leq 0 \\ -x^2 +1, & \text{if}\ 0 \leq x < 1\\ 0 & \text{otherwise} \end{cases} $

Also, is the general approach?

JOKKINATOR
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1 Answers1

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No.

If the distribution of $X$ is known and has PDF $f_X$ then for a suitable function $g:\mathbb R\to\mathbb R$ we find:$$\mathbb Eg(X)=\int g(x)f_X(x)dx\tag1$$

This equality carries the name "The law of the unconscious statistician" and also "Transfer Theorem" (see comment of Florian).

You can apply that here to find $\mathbb EX^2$ where function $g$ is obviously prescribed by $x\mapsto x^2$ and $f_X$ is the density described in your question.

Also have a look here.

For more info about $(1)$ you can take a look at (the answer on) this question.

drhab
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  • Maybe you should mention that this is called the Transfer Theorem (at least, we call it like this in France) as it is very useful in many cases. – Florian Ingels Apr 17 '19 at 12:46
  • @FlorianIngels I provided a link now if the OP wants more of the backgrounds. – drhab Apr 17 '19 at 12:52
  • Thank you both for you answers, I will look into this. My book makes it looks as if this is equivalent to substituting $x$ by $x^2$ @drhab – JOKKINATOR Apr 17 '19 at 13:19