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Calculate E|Y| where Y is the standard normal. (Hints: (1) Calculate the integrals for positive and negative values of Y separately. (2) The answer must be bigger than 0.)


Not sure how to approach this since no number were given. What am I supposed to be taking the integral of? I know that $E(X) = \int_0^1 x_iP(X=x_i)$ but I'm not too entirely sure how this would help me finding out the expectation of the absolute value? Is it just $E(X) = \int_0^1 |x_i|P(X=x_i)$ ?(I'm assuming not since the hint told me to integrate).

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    $\begingroup$ in general you have $E[g(X)] = \int_{-\infty}^{\infty} g(x) f(x) d x$ for any random variable $X$ with density function $f$. $\endgroup$
    – user251257
    Aug 29, 2016 at 22:50

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\begin{align*} E[|Y|] &= \int_{-\infty}^\infty |y| \dfrac{1}{\sqrt{2\pi}} e^{-y^2/2} \; dy \\ &= \int_{-\infty}^0 -y \dfrac{1}{\sqrt{2\pi}} e^{-y^2/2} \; dy + \int_0^\infty y \dfrac{1}{\sqrt{2\pi}} e^{-y^2/2} \; dy \\ &=\dots\\ &= \sqrt{2/\pi} \end{align*}

and others have given the general form of $E[g(Y)]$, which is how I got this.

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  • $\begingroup$ Is there a two dimensional analogue for this? $\endgroup$ Mar 27, 2019 at 18:34
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I don't know what you mean by "no number were given". The question is perfectly well specified. You do know what "standard normal" means?

For a continuous random variable $Y$ with density $f$ and some function $g$,

$$ E[g(Y)] = \int_{-\infty}^\infty g(y) f(y) \; dy$$

Here $g(y) = |y|$. As suggested in the hint, you should do $\int_{-\infty}^0$ and $\int_0^\infty$ separately (but by symmetry, the answers will be the same).

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