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I am trying to figure out if I made a mistake in the following proof - in particular I have been trying to verify the inequality (*) below. I have attached my proof but I have been getting some error messages in Mathematica that make me think something is wrong when I try to compare the left hand side of the inequality to the right hand side. The background of this problem is related to the study of moments of self similar processes and I have no idea if the estimates are valid besides from some crude numerical calculations for $N= 2,3,4$ and all $H_i = 1/2$. I have attached the proof for the case N =2.

1) Did I make a mistake in one of the estimates or applying fubini somewhere? In particular I am starting to worry about line (**) is breaking the integral up like this and then using $x_1 < x_2$ and $x_2 < x_1$ in the respective regions of integration to arrive at the estimate on the next line valid?

Let $H_1 \in (0,1), c$ a positive constant and let $\Phi(t) = \int_{t}^{\infty} e^{\frac{-x^2}{2}}dx$. Then we have the following inequality

(*) $\int_{1}^{\infty} \int_{1}^{\infty} e^{-\frac{ x_{1}^{2}}{2} -\frac{ x_{2}^{2}}{2} + \frac{1}{c} |x_1| |x_2|^{H_1} } dx_1 dx_2 \leq 2 \Phi(1) \int_{1}^{\infty} e^{ -\frac{ x_{2}^{2}}{2} + \frac{1}{c} |x_2| ^{1+H_1} }dx_2$

Proof:

$\int_{1}^{\infty} \int_{1}^{\infty} e^{-\frac{ x_{1}^{2}}{2} -\frac{ x_{2}^{2}}{2} + \frac{1}{c} |x_1| |x_2|^{H_1} } dx_1 dx_2$

(**) $ = \int_{1}^{\infty} \int_{1}^{x_2} e^{-\frac{ x_{1}^{2}}{2} -\frac{ x_{2}^{2}}{2} + \frac{1}{c} |x_1| |x_2|^{H_1} } dx_1 dx_2 + \int_{1}^{\infty} \int_{x_2}^{\infty} e^{-\frac{ x_{1}^{2}}{2} -\frac{ x_{2}^{2}}{2} + \frac{1}{c} |x_1| |x_2|^{H_1} } dx_1 dx_2 $

$ \leq \int_{1}^{\infty} \int_{1}^{x_2} e^{-\frac{ x_{1}^{2}}{2} -\frac{ x_{2}^{2}}{2} + \frac{1}{c} |x_2| ^{1+H_1} } dx_1 dx_2 + \int_{1}^{\infty} \int_{x_2}^{\infty} e^{-\frac{ x_{1}^{2}}{2} -\frac{ x_{2}^{2}}{2} + \frac{1}{c} |x_1|^{1+H_1} } dx_1 dx_2 $

$ = \int_{1}^{\infty} e^{ -\frac{ x_{2}^{2}}{2} + \frac{1}{c} |x_2| ^{1+H_1} } (\int_{1}^{x_2} e^{-\frac{ x_{1}^{2}}{2}}dx_1)dx_2 + \int_{1}^{\infty} e^{ -\frac{ x_{1}^{2}}{2} + \frac{1}{c} |x_1| ^{1+H_1} } (\int_{x_2}^{\infty} e^{-\frac{ x_{2}^{2}}{2}}dx_2)dx_1 $

$ \leq 2 \int_{1}^{\infty} e^{ -\frac{ x_{2}^{2}}{2} + \frac{1}{c} |x_2| ^{1+H_1} } (\int_{1}^{\infty} e^{-\frac{ x_{1}^{2}}{2}}dx_1)dx_2 = 2 \Phi(1) \int_{1}^{\infty} e^{ -\frac{ x_{2}^{2}}{2} + \frac{1}{c} |x_2| ^{1+H_1} }dx_2$

2)I have tried comparing the LHS to the right hand side for the case $N = 2, N= 4$ and all H_i = 1/2 but Mathematica gives me the following errors attached below. I am thinking I will just need to test the above inequality's for $N > 3$ in matlab but I am not sure what the easiest method would be. Would monte carlo integration apply to these type of integrals even though they are over an unbounded region? Any numerical suggestions/references are greatly appreciated

NIntegrate[ 2*(1/Sqrt[2*Pi])^2*Exp [1/2 (-x ^2 - y^2)] + 1/10*x*y^(1/2), {x, 0, Infinity}, {y, 0, Infinity}]

NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small. >>

NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 18 recursive bisections in x near {x,y} = {4.557658617658054*10^1348705,0.184661}. NIntegrate obtained 1.54175165043030515.954589770191005*^88741243 and 3.65322886863091515.954589770191005*^88741243 for the integral and error estimates. >>

0.*10^88741243

NIntegrate[ 2*(1/Sqrt[2*Pi])^4 Exp [ 1/2 (-x^2 - y^2 - z^2 - w^2) + 1/10 x Sqrt[y] z^(1/4)* w^(1/8)], {x, 0, [Infinity]}, {y, 0, [Infinity]}, {z, 0, [Infinity]}, {w, 0, [Infinity]}]

NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small. >>

NIntegrate::eincr: The global error of the strategy GlobalAdaptive has increased more than 2000 times. The global error is expected to decrease monotonically after a number of integrand evaluations. Suspect one of the following: the working precision is insufficient for the specified precision goal; the integrand is highly oscillatory or it is not a (piecewise) smooth function; or the true value of the integral is 0. Increasing the value of the GlobalAdaptive option MaxErrorIncreases might lead to a convergent numerical integration. NIntegrate obtained 0.1321276729737014and 4.5493840913708495*^-7 for the integral and error estimates. >>

0.132128

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1) Your proof of (*) seems OK to me; splitting up the integral as you do at (**) and then using $x_1 < x_2$ / $x_1 > x_2$ should be fine. (I didn't look at the Mathematica stuff.) –  joriki Mar 9 '11 at 12:05
    
thank you for your help –  user7980 Mar 11 '11 at 2:47

1 Answer 1

up vote 1 down vote accepted

There is an error in your computation in the change of order of integration in the second integral of the RHS. It should be $$ \int_1^\infty\Bigl(\int_{x_2}^\infty \exp(-\frac{x_1^2}{2}-\frac{x_2^2}{2}+\frac{x_1^{1+H_1}}{c})dx_1\Bigr)dx_2= $$

$$ \int_1^\infty\exp(-\frac{x_1^2}{2}+\frac{x_1^{1+H_1}}{c})\Bigl(\int_{1}^{x_1}\exp(-\frac{x_2^2}{2})dx_2\Bigr)dx_1. $$ As for the Mathematica calculation, there is also an error. The term

+1/10*x*y^(1/2)

should go inside the Exp .

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Thanks for the help I am unable to upvote your answer but I appreciate your comments nonetheless. –  user7980 Mar 10 '11 at 5:40

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