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Let us say that $P$ is a normal random variable having expected value $\mu$ and variance $\sigma^2$. I am asked to compute the expected value of the variable $Y = |P|$.

Could someone explain?

cardinal
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Probabilityman
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  • You need to know more that the information given. Are you sure it wasn't $y = |p|^2$? – cardinal Jan 22 '12 at 21:51
  • Jensen's inequality gives you a simple upper bound since $$\mathbb E Y = \mathbb E \sqrt{P^2} \leq \sqrt{\mathbb E P^2} = \sqrt{\mu^2 + \sigma^2} > .$$ – cardinal Jan 22 '12 at 21:54
  • I am sure about the question. it is to compute the expected value of variable Y=|P|. – Probabilityman Jan 22 '12 at 21:55
  • Do you know something about the distribution of $P$? Perhaps it is normal? Or something else? Without further information a definitive answer is not possible. (Consider for example, if $\mathbb P(P=1) = \mathbb P(P=-1) = 1/2$ vs. the case where $P$ is standard normal. Both have mean zero and variance 1, but $\mathbb E |P|$ is different in the two cases.) – cardinal Jan 22 '12 at 21:57
  • P is a normal random variable – Probabilityman Jan 22 '12 at 21:59
  • Related: http://math.stackexchange.com/questions/67561/the-expectation-of-the-half-normal-distribution – cardinal Jan 22 '12 at 22:02
  • I went through the similar question you directed me to. so here's what I think and what i would like to further clarify. – Probabilityman Jan 22 '12 at 22:17
  • okay before i proceed will this be a discreet or continuous random variable? – Probabilityman Jan 22 '12 at 22:20

2 Answers2

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Hint:

By definition,

$$E(|P|)=\int_{-\infty}^\infty |x| P(x)dx=\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^\infty |x| e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx$$

You can divide the integral into $$\int_{-\infty}^\infty=\int_{-\infty}^0+\int_0^\infty$$ Now you can calculate these integrals (hint: what is $|x|$ for $x<0$? and for $x>0$?)

yohBS
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    A hint to add to yohBS's answer. Do the following seemingly mindless exercise. Differentiate $e^{-(x-\mu)^2/2\sigma^2}$ and stare very hard at the result and at the integrands above. – Dilip Sarwate Jan 22 '12 at 22:33
  • @DilipSarwate I get the answer as - (of the same value) as in -e^(-(x−μ)2/2σ2) – Probabilityman Jan 22 '12 at 23:22
  • Is thsi the correct approach. I used |x| for x<0 as -x and |x| for x>0 as +x. also my final answer what i obtained is: – Probabilityman Jan 23 '12 at 00:23
  • 1/(sqrt(2pisigma sq) * (2e^-mu sq + root(pi)mulog(e))/(4sigma square log(e)) – Probabilityman Jan 23 '12 at 00:24
  • @cardinal if you could provide me your email id, i could email you a snap shot of the whole integration process. i am unable to type out the entire thing, i did it on paper – Probabilityman Jan 23 '12 at 00:26
  • For the final answer, you can look in wikipedia - http://en.wikipedia.org/wiki/Folded_normal_distribution – yohBS Jan 23 '12 at 07:00
  • Is my approach correct in using -x in the first integral and x in the second.. Also shouldnt the X in the exp(x-mu/sigma) be alternatively substituted depending on the value of X in the integral. would it be right to replace exp(x-mu/sigma) by z and then integrate to obtain the final integral? – Probabilityman Jan 23 '12 at 14:52
  • about your first question - yes. I don't understand the second one. – yohBS Jan 23 '12 at 19:00
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Y has folded normal distribution, in the following link you can find its expected value using the expected value and variance of P. http://en.wikipedia.org/wiki/Folded_normal_distribution

May
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