Asymptotic expansion of integral with standard normal distribution cdf

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I am looking for the asymptotic expansion of the following expression as $taurightarrow0$ :

$F(tau) = int_0^taue^-rumathcalNleft(fraclogleft(fracbarxleft(tau-uright)x+aright)-mu usigmasqrturight)du$, where :

$r>0, sigma>0, mu>0, a>0, mathcalN$ the c.d.f of a standard normal distribution and $barx(tau)sim_0 (x+a)(1-sigmasqrt-taulntau)$

First, note that :

$F(tau)=tauint_0^1e^-rtau umathcalNleft(fraclogleft(fracbarxleft(tauleft(1-uright)right)x+aright)-mu utausigmasqrttau uright)d u$

Now, let v:=1-u and :

$logleft(fracbarxleft(vtauright)x+aright) =logleft(fracx+aleft(1-sigmasqrt-tau vlogtau vright)x+a+oleft(-sqrt-tau vlogtau vright)right)
=-sigmasqrt-tau vlogtau v+oleft(sqrt-tau vlogtau vright)$

Hence :

$fraclogleft(fracbarxleft(tau vright)x+aright)-mu utausigmasqrttau u=-sqrt-fracvulogtau v+oleft(-sqrt-fracvulogtau vright)=-sqrt-fracvulogtau+oleft(-sqrt-logtauright)$

Now, recall that one has :

$mathcalNleft(xright)underset-inftysim-expleft(-fracx^22right)frac1sqrt2pix$

Finally :

$Fleft(tauright)=tauint_0^1left(1-rtau u+oleft(tau uright)right)fracexpleft[frac1-u2uleft(logleft(tauright)+oleft(logleft(tauright)right)right)right]sqrt2pileft(-sqrt-fracvulogtau+oleft(-sqrt-logtauright)right)du$

$F(tau)=tauint_0^1fracexpleft[frac1-u2uleft(logleft(tauright)+oleft(logleft(tauright)right)right)right]sqrt2pileft(-sqrt-fracvulogtau+oleft(-sqrt-logtauright)right)du+Oleft(tau^2right)$

Computing the integral yields :

$sqrt-fractaulogtaufracsqrt2pisqrt-taulogtau+pitexterfcleft[frac-sqrtlogtausqrt2right]left(1+logtauright)2sqrt2pi$ and using a Taylor expansion, we finally obtain :

$F(tau)sim_0 sqrtfrac-1logtaufracleft(3+2logtauleft(1+logtauright)right)tau2left(-logtauright)^frac52+Oleft(tau^2right)$

Do you agree with the following ?

real-analysis limits taylor-expansion special-functions

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asked Jul 18 at 11:20

flo3299

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I am looking for the asymptotic expansion of the following expression as $taurightarrow0$ :

$F(tau) = int_0^taue^-rumathcalNleft(fraclogleft(fracbarxleft(tau-uright)x+aright)-mu usigmasqrturight)du$, where :

$r>0, sigma>0, mu>0, a>0, mathcalN$ the c.d.f of a standard normal distribution and $barx(tau)sim_0 (x+a)(1-sigmasqrt-taulntau)$

First, note that :

$F(tau)=tauint_0^1e^-rtau umathcalNleft(fraclogleft(fracbarxleft(tauleft(1-uright)right)x+aright)-mu utausigmasqrttau uright)d u$

Now, let v:=1-u and :

$logleft(fracbarxleft(vtauright)x+aright) =logleft(fracx+aleft(1-sigmasqrt-tau vlogtau vright)x+a+oleft(-sqrt-tau vlogtau vright)right)
=-sigmasqrt-tau vlogtau v+oleft(sqrt-tau vlogtau vright)$

Hence :

$fraclogleft(fracbarxleft(tau vright)x+aright)-mu utausigmasqrttau u=-sqrt-fracvulogtau v+oleft(-sqrt-fracvulogtau vright)=-sqrt-fracvulogtau+oleft(-sqrt-logtauright)$

Now, recall that one has :

$mathcalNleft(xright)underset-inftysim-expleft(-fracx^22right)frac1sqrt2pix$

Finally :

$Fleft(tauright)=tauint_0^1left(1-rtau u+oleft(tau uright)right)fracexpleft[frac1-u2uleft(logleft(tauright)+oleft(logleft(tauright)right)right)right]sqrt2pileft(-sqrt-fracvulogtau+oleft(-sqrt-logtauright)right)du$

$F(tau)=tauint_0^1fracexpleft[frac1-u2uleft(logleft(tauright)+oleft(logleft(tauright)right)right)right]sqrt2pileft(-sqrt-fracvulogtau+oleft(-sqrt-logtauright)right)du+Oleft(tau^2right)$

Computing the integral yields :

$sqrt-fractaulogtaufracsqrt2pisqrt-taulogtau+pitexterfcleft[frac-sqrtlogtausqrt2right]left(1+logtauright)2sqrt2pi$ and using a Taylor expansion, we finally obtain :

$F(tau)sim_0 sqrtfrac-1logtaufracleft(3+2logtauleft(1+logtauright)right)tau2left(-logtauright)^frac52+Oleft(tau^2right)$

Do you agree with the following ?

real-analysis limits taylor-expansion special-functions

share|cite|improve this question

asked Jul 18 at 11:20

flo3299

164

add a comment | 

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I am looking for the asymptotic expansion of the following expression as $taurightarrow0$ :

$F(tau) = int_0^taue^-rumathcalNleft(fraclogleft(fracbarxleft(tau-uright)x+aright)-mu usigmasqrturight)du$, where :

$r>0, sigma>0, mu>0, a>0, mathcalN$ the c.d.f of a standard normal distribution and $barx(tau)sim_0 (x+a)(1-sigmasqrt-taulntau)$

First, note that :

$F(tau)=tauint_0^1e^-rtau umathcalNleft(fraclogleft(fracbarxleft(tauleft(1-uright)right)x+aright)-mu utausigmasqrttau uright)d u$

Now, let v:=1-u and :

$logleft(fracbarxleft(vtauright)x+aright) =logleft(fracx+aleft(1-sigmasqrt-tau vlogtau vright)x+a+oleft(-sqrt-tau vlogtau vright)right)
=-sigmasqrt-tau vlogtau v+oleft(sqrt-tau vlogtau vright)$

Hence :

$fraclogleft(fracbarxleft(tau vright)x+aright)-mu utausigmasqrttau u=-sqrt-fracvulogtau v+oleft(-sqrt-fracvulogtau vright)=-sqrt-fracvulogtau+oleft(-sqrt-logtauright)$

Now, recall that one has :

$mathcalNleft(xright)underset-inftysim-expleft(-fracx^22right)frac1sqrt2pix$

Finally :

$Fleft(tauright)=tauint_0^1left(1-rtau u+oleft(tau uright)right)fracexpleft[frac1-u2uleft(logleft(tauright)+oleft(logleft(tauright)right)right)right]sqrt2pileft(-sqrt-fracvulogtau+oleft(-sqrt-logtauright)right)du$

$F(tau)=tauint_0^1fracexpleft[frac1-u2uleft(logleft(tauright)+oleft(logleft(tauright)right)right)right]sqrt2pileft(-sqrt-fracvulogtau+oleft(-sqrt-logtauright)right)du+Oleft(tau^2right)$

Computing the integral yields :

$sqrt-fractaulogtaufracsqrt2pisqrt-taulogtau+pitexterfcleft[frac-sqrtlogtausqrt2right]left(1+logtauright)2sqrt2pi$ and using a Taylor expansion, we finally obtain :

$F(tau)sim_0 sqrtfrac-1logtaufracleft(3+2logtauleft(1+logtauright)right)tau2left(-logtauright)^frac52+Oleft(tau^2right)$

Do you agree with the following ?

real-analysis limits taylor-expansion special-functions

share|cite|improve this question

asked Jul 18 at 11:20

flo3299

164

I am looking for the asymptotic expansion of the following expression as $taurightarrow0$ :

$F(tau) = int_0^taue^-rumathcalNleft(fraclogleft(fracbarxleft(tau-uright)x+aright)-mu usigmasqrturight)du$, where :

$r>0, sigma>0, mu>0, a>0, mathcalN$ the c.d.f of a standard normal distribution and $barx(tau)sim_0 (x+a)(1-sigmasqrt-taulntau)$

First, note that :

$F(tau)=tauint_0^1e^-rtau umathcalNleft(fraclogleft(fracbarxleft(tauleft(1-uright)right)x+aright)-mu utausigmasqrttau uright)d u$

Now, let v:=1-u and :

$logleft(fracbarxleft(vtauright)x+aright) =logleft(fracx+aleft(1-sigmasqrt-tau vlogtau vright)x+a+oleft(-sqrt-tau vlogtau vright)right)
=-sigmasqrt-tau vlogtau v+oleft(sqrt-tau vlogtau vright)$

Hence :

$fraclogleft(fracbarxleft(tau vright)x+aright)-mu utausigmasqrttau u=-sqrt-fracvulogtau v+oleft(-sqrt-fracvulogtau vright)=-sqrt-fracvulogtau+oleft(-sqrt-logtauright)$

Now, recall that one has :

$mathcalNleft(xright)underset-inftysim-expleft(-fracx^22right)frac1sqrt2pix$

Finally :

$Fleft(tauright)=tauint_0^1left(1-rtau u+oleft(tau uright)right)fracexpleft[frac1-u2uleft(logleft(tauright)+oleft(logleft(tauright)right)right)right]sqrt2pileft(-sqrt-fracvulogtau+oleft(-sqrt-logtauright)right)du$

$F(tau)=tauint_0^1fracexpleft[frac1-u2uleft(logleft(tauright)+oleft(logleft(tauright)right)right)right]sqrt2pileft(-sqrt-fracvulogtau+oleft(-sqrt-logtauright)right)du+Oleft(tau^2right)$

Computing the integral yields :

$sqrt-fractaulogtaufracsqrt2pisqrt-taulogtau+pitexterfcleft[frac-sqrtlogtausqrt2right]left(1+logtauright)2sqrt2pi$ and using a Taylor expansion, we finally obtain :

$F(tau)sim_0 sqrtfrac-1logtaufracleft(3+2logtauleft(1+logtauright)right)tau2left(-logtauright)^frac52+Oleft(tau^2right)$

Do you agree with the following ?

real-analysis limits taylor-expansion special-functions

share|cite|improve this question

asked Jul 18 at 11:20

flo3299

164

share|cite|improve this question

share|cite|improve this question

asked Jul 18 at 11:20

flo3299

164

asked Jul 18 at 11:20

flo3299

164

asked Jul 18 at 11:20

flo3299

164

164

add a comment | 

add a comment | 

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