expected value of a normally distributed variable

**minivan15** · June 8th 2009, 07:19 PM

Hey...I've been thinking about this problem all day, and can't seem to come up with an exact answer.

consider a standard normally distributed random variable, say X. What is the expected value of X, given it is between two values (say 1 and -2)?

**mr fantastic** · June 8th 2009, 07:23 PM

Originally Posted by minivan15

Hey...I've been thinking about this problem all day, and can't seem to come up with an exact answer.

consider a standard normally distributed random variable, say X. What is the expected value of X, given it is between two values (say 1 and -2)?

Apply the ideas found in this link: Conditional Expectation

**Random Variable** · June 8th 2009, 08:52 PM

In general, let X be any continuous random variable.

Then $E[X|a<X<b] = \int^{b}_{a} \frac {x f(x)}{P(a<x<b)} \ dx$

**matheagle** · June 8th 2009, 10:05 PM

The X is a rv, not a variable of integration...

Originally Posted by Random Variable

In general, let X be any continuous random variable.

Then $E[X|a<X<b] = {\int^{b}_{a} x f_X(x)dx\over P(a<X<b)}$

**Random Variable** · June 8th 2009, 10:24 PM

I stated that X is a random variable. And of course I meant to capitalize the X in the denominator. P(a<x<b) has no meaning. And there is nothing wrong with the way I wrote it. The denominator is a number.

I know you think I'm incredibly stupid, but give me some credit. Sheesh.

**matheagle** · June 8th 2009, 10:31 PM

The comment was not for you, but for the poster.
The probability could be in or out of the integral, but with it inside and being lower case as with the x in the numerator it is misleading.
It could easily be interpreted as part of the integrand.

And I do not think you're stupid at all.

**Random Variable** · June 8th 2009, 11:16 PM

Originally Posted by matheagle

The comment was not for you, but for the poster.
The probability could be in or out of the integral, but with it inside and being lower case as with the x in the numerator it is misleading.
It could easily be interpreted as part of the integrand.

And I do not think you're stupid at all.

I apologize. I thought you were talking directly to me.

But yes, it could cause confusion.

**minivan15** · June 9th 2009, 12:09 PM

so, just to make sure I have this...

the expected value of a std normal variable X, given X is between -2 and 1, is about -.229 (can someone verify this answer?)

**Random Variable** · June 9th 2009, 12:48 PM

Originally Posted by minivan15

so, just to make sure I have this...

the expected value of a std normal variable X, given X is between -2 and 1, is about -.229 (can someone verify this answer?)

$E[X|-2<X<1] = \frac {\frac {1}{\sqrt{2 \pi}} \int^{1}_{-2} x e^{- \frac {x^{2}}{2}} \ dx}{ \frac {1}{\sqrt{2 \pi}} \int^{1}_{-2}e^{- \frac {x^{2}}{2}}\ dx} \approx -0.2296$

yep

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