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I am reading a book on antithetic sampling.It is said that the idea of antithetic sampling can be applied when it is possible to find transformation of $X$ that leave its measure unchanged (for example, if $X$ is Gaussian, then $-X$ is Gaussian as well).Suppose that we want to calculate $$ I=\int_0^1{g(x)dx}=E[g(x)] $$ with $X$~$U(0,1)$, and $U$ denotes uniform distribution.The transformation $x\rightarrow 1-x$ leaves the measure unchanged (i.e,$1-X$~$U(0,1)$), and $I$ can be rewritten as $$ I=\frac{1}{2}\int_0^1{(g(x)+g(1-x)})dx. $$ My question is:

Is the second integration only apply for uniformly distributed variables? Or it can be used for variables with any distribution? Thank you!

Jackie
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  • For this question, I got the answer from the author of the book.The second integration only apply for uniformly distributed variables. – Jackie Oct 20 '14 at 17:02

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For any random variable $X$ which is symmetric with respect to $\frac{1}{2}$, $X$ and $1-X$ have the same distribution.

Petite Etincelle
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  • So if $X$ is symmetric with respect to $\frac{2}{5}$, then $\frac{1}{2}$ of the second integration should be substituted by $\frac{2}{5}$? Thanks again! – Jackie Oct 19 '14 at 05:48
  • @Jackie if symmetric with respect to $\frac{2}{5}$, you will have $Eg(X) = \frac{1}{2}(Eg(X) + Eg(\frac{4}{5}-X))$. Don't forget the density function when expressing expectation as integral. In you second integration the density happens to be $1$ – Petite Etincelle Oct 19 '14 at 06:54
  • should the righthand side of the equation be $\frac{1}{2}(Eg(X)+Eg(\frac{2}{5} - X))$? And,the integration should be $\frac{1}{2} \int_0^1 {(g(x)+g(\frac{2}{5} -x)) f(x) dx}$? Since the limits of $X$ is $[0,1]$,so $f(x)=1$? :-) – Jackie Oct 20 '14 at 06:12
  • @Jackie When symmetric with respect to $a$, when $X$ and $2a-X$ are of the same distribution. It's ok to take limit of $f(x)$ under some regularity conditions, but if the limit of $X$ is $[0,1]$, how can it be symmetric w.r.t $\frac{2}{5}$? – Petite Etincelle Oct 20 '14 at 07:14
  • so $f(x)=\frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} x^2}$ in case of $X$ with normal distribution? – Jackie Oct 20 '14 at 07:48
  • Thanks so much for your comments.I have poor knowledge on 'symmetry' analysis and need to replenish my knowledge. Could you please recommend any books? – Jackie Oct 20 '14 at 07:53
  • @Jackie the normal distribution is symmetric with respect to $0$. I think you just need to draw a graph to see in the plane $R^2$, if two points $(x_1,y)$ and $(x_2,y)$ are symmetric w.r.t the vertical line $x=a$, how to write $x_2$ in terms of $x_1$ and $a$?(answer: $x_2 = 2a - x_1$) – Petite Etincelle Oct 20 '14 at 09:00
  • I know how to get $x_1$ and $x_2$ given $a$.I don't know how to get the density. – Jackie Oct 20 '14 at 09:17
  • @Jackie you decide what kind of random variable $X$ is, then you know its density, no? – Petite Etincelle Oct 20 '14 at 10:00