Just how do I compute anticipated value of partial regular circulation?

Intend you have a regular circulation with mean= 0, and also stdev= 1. So the anticipated value is 0.

Currently intend you restrict the end results, such that no values can be listed below 0. So 50% of values currently equivalent 0, and also remainder of circulation is still regular. Running 1000000 tests, I bring out an anticipated value of.4

My inquiry is just how can I get this anticipated value via estimation?

Many thanks

2019-05-04 06:29:48
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Answers: 1

The regular circulation has thickness function $f(x)=\frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}$ ; your new circulation has that thickness function on the favorable reals, $P(0)=\frac{1}{2}$, and also $P(x)=0$ for the adverse reals. The anticipated value is $0\cdot\frac{1}{2}+\int_{0}^{\infty}x\cdot f(x)dx=\frac{1}{\sqrt{2\pi}}\approx0.398942$.

modify : If you were to remove at $x=c$ (appointing all the chance from listed below c to c itself) as opposed to $x=0$, your thickness function would certainly be $f(x)=\frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}$ for $x>c$, $P(c)=\int_{-\infty}^{c}\frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}dx$, and also $P(x)=0$ for $x<c$, so the anticipated value is $c\cdot P(c) + \int_{c}^{\infty}x\cdot \frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}dx$.

modify 2 : note that the backer on e in all of the above is $-\frac{x^2}{2}$ (the backer 2 on the x is, in the existing TeX making, located and also sized such as to be rather unclear)

modify 3 : my description inaccurately combined chance thickness features and also actual chances-- this was only a concern of terms and also the analytic outcomes still stand, yet I have actually tried to make clear the language above.

2019-05-08 03:44:33