Laplace transform of a normal distribution












1












$begingroup$


Let $X$ be a normal distribution with mean $0$ and variance $sigma^2$. Then the Laplace transform of $X$ should be



$int_0^infty e^{-tx} frac{1}{sqrt{2pisigma^2}}exp{-frac{x^2}{2sigma^2}}dx$



Then I try to do integral on



$int_0^infty exp{-frac{x^2}{2sigma^2}-tx}dx=int_0^infty exp{-frac{1}{2sigma^2}[x^2+2sigma^2tx]}dx$



finishing the square we can get



$int_0^infty exp{-frac{1}{2sigma^2}(x+sigma^2 t)^2}dx$



Then I replace $x+sigma^2 t$ by $u$



$int_{sigma^2t}^infty exp{-frac{1}{2sigma^2}u^2}du$



Then I am stuck at this step.






probability-theory probability-distributions normal-distribution laplace-transform







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$endgroup$












  • $begingroup$
    This is as far as it goes: the answer is in terms of the cumulative normal distribution function.
    $endgroup$
    – uniquesolution
    Sep 18 '17 at 15:53












  • $begingroup$
    finishing the square yields another gaussian
    $endgroup$
    – alain
    Jan 22 '18 at 10:51
















1












$begingroup$


Let $X$ be a normal distribution with mean $0$ and variance $sigma^2$. Then the Laplace transform of $X$ should be



$int_0^infty e^{-tx} frac{1}{sqrt{2pisigma^2}}exp{-frac{x^2}{2sigma^2}}dx$



Then I try to do integral on



$int_0^infty exp{-frac{x^2}{2sigma^2}-tx}dx=int_0^infty exp{-frac{1}{2sigma^2}[x^2+2sigma^2tx]}dx$



finishing the square we can get



$int_0^infty exp{-frac{1}{2sigma^2}(x+sigma^2 t)^2}dx$



Then I replace $x+sigma^2 t$ by $u$



$int_{sigma^2t}^infty exp{-frac{1}{2sigma^2}u^2}du$



Then I am stuck at this step.






probability-theory probability-distributions normal-distribution laplace-transform







share|cite|improve this question









$endgroup$












  • $begingroup$
    This is as far as it goes: the answer is in terms of the cumulative normal distribution function.
    $endgroup$
    – uniquesolution
    Sep 18 '17 at 15:53












  • $begingroup$
    finishing the square yields another gaussian
    $endgroup$
    – alain
    Jan 22 '18 at 10:51














1












1








1


1



$begingroup$


Let $X$ be a normal distribution with mean $0$ and variance $sigma^2$. Then the Laplace transform of $X$ should be



$int_0^infty e^{-tx} frac{1}{sqrt{2pisigma^2}}exp{-frac{x^2}{2sigma^2}}dx$



Then I try to do integral on



$int_0^infty exp{-frac{x^2}{2sigma^2}-tx}dx=int_0^infty exp{-frac{1}{2sigma^2}[x^2+2sigma^2tx]}dx$



finishing the square we can get



$int_0^infty exp{-frac{1}{2sigma^2}(x+sigma^2 t)^2}dx$



Then I replace $x+sigma^2 t$ by $u$



$int_{sigma^2t}^infty exp{-frac{1}{2sigma^2}u^2}du$



Then I am stuck at this step.






probability-theory probability-distributions normal-distribution laplace-transform







share|cite|improve this question









$endgroup$




Let $X$ be a normal distribution with mean $0$ and variance $sigma^2$. Then the Laplace transform of $X$ should be



$int_0^infty e^{-tx} frac{1}{sqrt{2pisigma^2}}exp{-frac{x^2}{2sigma^2}}dx$



Then I try to do integral on



$int_0^infty exp{-frac{x^2}{2sigma^2}-tx}dx=int_0^infty exp{-frac{1}{2sigma^2}[x^2+2sigma^2tx]}dx$



finishing the square we can get



$int_0^infty exp{-frac{1}{2sigma^2}(x+sigma^2 t)^2}dx$



Then I replace $x+sigma^2 t$ by $u$



$int_{sigma^2t}^infty exp{-frac{1}{2sigma^2}u^2}du$



Then I am stuck at this step.






probability-theory probability-distributions normal-distribution laplace-transform




probability-theory probability-distributions normal-distribution laplace-transform






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 18 '17 at 15:42









Kenneth.KKenneth.K

499316




499316












  • $begingroup$
    This is as far as it goes: the answer is in terms of the cumulative normal distribution function.
    $endgroup$
    – uniquesolution
    Sep 18 '17 at 15:53












  • $begingroup$
    finishing the square yields another gaussian
    $endgroup$
    – alain
    Jan 22 '18 at 10:51


















  • $begingroup$
    This is as far as it goes: the answer is in terms of the cumulative normal distribution function.
    $endgroup$
    – uniquesolution
    Sep 18 '17 at 15:53












  • $begingroup$
    finishing the square yields another gaussian
    $endgroup$
    – alain
    Jan 22 '18 at 10:51
















$begingroup$
This is as far as it goes: the answer is in terms of the cumulative normal distribution function.
$endgroup$
– uniquesolution
Sep 18 '17 at 15:53






$begingroup$
This is as far as it goes: the answer is in terms of the cumulative normal distribution function.
$endgroup$
– uniquesolution
Sep 18 '17 at 15:53














$begingroup$
finishing the square yields another gaussian
$endgroup$
– alain
Jan 22 '18 at 10:51




$begingroup$
finishing the square yields another gaussian
$endgroup$
– alain
Jan 22 '18 at 10:51










1 Answer
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$begingroup$

Well, in the general case you're looking at:



$$mathscr{I}_{spacetext{n}}left(text{m}space,spacetext{s}right):=int_0^text{m}frac{text{n}}{expleft(text{s}cdot x+picdottext{n}^2cdot x^2right)}spacetext{d}xtag1$$



Substitute:



$$text{u}:=frac{2cdotpicdottext{n}^2cdot x+text{s}}{2cdotsqrt{pi}cdottext{n}}tag2$$



So, for the integral we get:



$$mathscr{I}_{spacetext{n}}left(text{m}space,spacetext{s}right):=text{n}cdotfrac{expleft(frac{text{s}^2}{4cdotpicdottext{n}^2}right)}{2cdottext{n}}int_{frac{text{s}}{2cdotsqrt{pi}cdottext{n}}}^{frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}}frac{2cdotexpleft(-text{u}^2right)}{sqrt{pi}}spacetext{d}text{u}tag2$$



This is a special integral:



$$int_{frac{text{s}}{2cdotsqrt{pi}cdottext{n}}}^{frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}}frac{2cdotexpleft(-text{u}^2right)}{sqrt{pi}}spacetext{d}text{u}=left[text{erf}left(text{u}right)right]_{frac{text{s}}{2cdotsqrt{pi}cdottext{n}}}^{frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}}=$$
$$text{erf}left(frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}right)-text{erf}left(frac{text{s}}{2cdotsqrt{pi}cdottext{n}}right)tag3$$



So, we end up with:



$$mathscr{I}_{spacetext{n}}left(text{m}space,spacetext{s}right)=text{n}cdotfrac{expleft(frac{text{s}^2}{4cdotpicdottext{n}^2}right)}{2cdottext{n}}cdotleft{text{erf}left(frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}right)-text{erf}left(frac{text{s}}{2cdotsqrt{pi}cdottext{n}}right)right}tag4$$




Now, your task is to simplify and to take:



$$lim_{text{m}toinfty}mathscr{I}_{spacetext{n}}left(text{m}space,spacetext{s}right)tag5$$





Try to prove:



$$lim_{text{m}toinfty}text{erf}left(frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}right)=1tag6$$






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    0












    $begingroup$

    Well, in the general case you're looking at:



    $$mathscr{I}_{spacetext{n}}left(text{m}space,spacetext{s}right):=int_0^text{m}frac{text{n}}{expleft(text{s}cdot x+picdottext{n}^2cdot x^2right)}spacetext{d}xtag1$$



    Substitute:



    $$text{u}:=frac{2cdotpicdottext{n}^2cdot x+text{s}}{2cdotsqrt{pi}cdottext{n}}tag2$$



    So, for the integral we get:



    $$mathscr{I}_{spacetext{n}}left(text{m}space,spacetext{s}right):=text{n}cdotfrac{expleft(frac{text{s}^2}{4cdotpicdottext{n}^2}right)}{2cdottext{n}}int_{frac{text{s}}{2cdotsqrt{pi}cdottext{n}}}^{frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}}frac{2cdotexpleft(-text{u}^2right)}{sqrt{pi}}spacetext{d}text{u}tag2$$



    This is a special integral:



    $$int_{frac{text{s}}{2cdotsqrt{pi}cdottext{n}}}^{frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}}frac{2cdotexpleft(-text{u}^2right)}{sqrt{pi}}spacetext{d}text{u}=left[text{erf}left(text{u}right)right]_{frac{text{s}}{2cdotsqrt{pi}cdottext{n}}}^{frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}}=$$
    $$text{erf}left(frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}right)-text{erf}left(frac{text{s}}{2cdotsqrt{pi}cdottext{n}}right)tag3$$



    So, we end up with:



    $$mathscr{I}_{spacetext{n}}left(text{m}space,spacetext{s}right)=text{n}cdotfrac{expleft(frac{text{s}^2}{4cdotpicdottext{n}^2}right)}{2cdottext{n}}cdotleft{text{erf}left(frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}right)-text{erf}left(frac{text{s}}{2cdotsqrt{pi}cdottext{n}}right)right}tag4$$




    Now, your task is to simplify and to take:



    $$lim_{text{m}toinfty}mathscr{I}_{spacetext{n}}left(text{m}space,spacetext{s}right)tag5$$





    Try to prove:



    $$lim_{text{m}toinfty}text{erf}left(frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}right)=1tag6$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Well, in the general case you're looking at:



      $$mathscr{I}_{spacetext{n}}left(text{m}space,spacetext{s}right):=int_0^text{m}frac{text{n}}{expleft(text{s}cdot x+picdottext{n}^2cdot x^2right)}spacetext{d}xtag1$$



      Substitute:



      $$text{u}:=frac{2cdotpicdottext{n}^2cdot x+text{s}}{2cdotsqrt{pi}cdottext{n}}tag2$$



      So, for the integral we get:



      $$mathscr{I}_{spacetext{n}}left(text{m}space,spacetext{s}right):=text{n}cdotfrac{expleft(frac{text{s}^2}{4cdotpicdottext{n}^2}right)}{2cdottext{n}}int_{frac{text{s}}{2cdotsqrt{pi}cdottext{n}}}^{frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}}frac{2cdotexpleft(-text{u}^2right)}{sqrt{pi}}spacetext{d}text{u}tag2$$



      This is a special integral:



      $$int_{frac{text{s}}{2cdotsqrt{pi}cdottext{n}}}^{frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}}frac{2cdotexpleft(-text{u}^2right)}{sqrt{pi}}spacetext{d}text{u}=left[text{erf}left(text{u}right)right]_{frac{text{s}}{2cdotsqrt{pi}cdottext{n}}}^{frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}}=$$
      $$text{erf}left(frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}right)-text{erf}left(frac{text{s}}{2cdotsqrt{pi}cdottext{n}}right)tag3$$



      So, we end up with:



      $$mathscr{I}_{spacetext{n}}left(text{m}space,spacetext{s}right)=text{n}cdotfrac{expleft(frac{text{s}^2}{4cdotpicdottext{n}^2}right)}{2cdottext{n}}cdotleft{text{erf}left(frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}right)-text{erf}left(frac{text{s}}{2cdotsqrt{pi}cdottext{n}}right)right}tag4$$




      Now, your task is to simplify and to take:



      $$lim_{text{m}toinfty}mathscr{I}_{spacetext{n}}left(text{m}space,spacetext{s}right)tag5$$





      Try to prove:



      $$lim_{text{m}toinfty}text{erf}left(frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}right)=1tag6$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Well, in the general case you're looking at:



        $$mathscr{I}_{spacetext{n}}left(text{m}space,spacetext{s}right):=int_0^text{m}frac{text{n}}{expleft(text{s}cdot x+picdottext{n}^2cdot x^2right)}spacetext{d}xtag1$$



        Substitute:



        $$text{u}:=frac{2cdotpicdottext{n}^2cdot x+text{s}}{2cdotsqrt{pi}cdottext{n}}tag2$$



        So, for the integral we get:



        $$mathscr{I}_{spacetext{n}}left(text{m}space,spacetext{s}right):=text{n}cdotfrac{expleft(frac{text{s}^2}{4cdotpicdottext{n}^2}right)}{2cdottext{n}}int_{frac{text{s}}{2cdotsqrt{pi}cdottext{n}}}^{frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}}frac{2cdotexpleft(-text{u}^2right)}{sqrt{pi}}spacetext{d}text{u}tag2$$



        This is a special integral:



        $$int_{frac{text{s}}{2cdotsqrt{pi}cdottext{n}}}^{frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}}frac{2cdotexpleft(-text{u}^2right)}{sqrt{pi}}spacetext{d}text{u}=left[text{erf}left(text{u}right)right]_{frac{text{s}}{2cdotsqrt{pi}cdottext{n}}}^{frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}}=$$
        $$text{erf}left(frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}right)-text{erf}left(frac{text{s}}{2cdotsqrt{pi}cdottext{n}}right)tag3$$



        So, we end up with:



        $$mathscr{I}_{spacetext{n}}left(text{m}space,spacetext{s}right)=text{n}cdotfrac{expleft(frac{text{s}^2}{4cdotpicdottext{n}^2}right)}{2cdottext{n}}cdotleft{text{erf}left(frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}right)-text{erf}left(frac{text{s}}{2cdotsqrt{pi}cdottext{n}}right)right}tag4$$




        Now, your task is to simplify and to take:



        $$lim_{text{m}toinfty}mathscr{I}_{spacetext{n}}left(text{m}space,spacetext{s}right)tag5$$





        Try to prove:



        $$lim_{text{m}toinfty}text{erf}left(frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}right)=1tag6$$






        share|cite|improve this answer









        $endgroup$



        Well, in the general case you're looking at:



        $$mathscr{I}_{spacetext{n}}left(text{m}space,spacetext{s}right):=int_0^text{m}frac{text{n}}{expleft(text{s}cdot x+picdottext{n}^2cdot x^2right)}spacetext{d}xtag1$$



        Substitute:



        $$text{u}:=frac{2cdotpicdottext{n}^2cdot x+text{s}}{2cdotsqrt{pi}cdottext{n}}tag2$$



        So, for the integral we get:



        $$mathscr{I}_{spacetext{n}}left(text{m}space,spacetext{s}right):=text{n}cdotfrac{expleft(frac{text{s}^2}{4cdotpicdottext{n}^2}right)}{2cdottext{n}}int_{frac{text{s}}{2cdotsqrt{pi}cdottext{n}}}^{frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}}frac{2cdotexpleft(-text{u}^2right)}{sqrt{pi}}spacetext{d}text{u}tag2$$



        This is a special integral:



        $$int_{frac{text{s}}{2cdotsqrt{pi}cdottext{n}}}^{frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}}frac{2cdotexpleft(-text{u}^2right)}{sqrt{pi}}spacetext{d}text{u}=left[text{erf}left(text{u}right)right]_{frac{text{s}}{2cdotsqrt{pi}cdottext{n}}}^{frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}}=$$
        $$text{erf}left(frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}right)-text{erf}left(frac{text{s}}{2cdotsqrt{pi}cdottext{n}}right)tag3$$



        So, we end up with:



        $$mathscr{I}_{spacetext{n}}left(text{m}space,spacetext{s}right)=text{n}cdotfrac{expleft(frac{text{s}^2}{4cdotpicdottext{n}^2}right)}{2cdottext{n}}cdotleft{text{erf}left(frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}right)-text{erf}left(frac{text{s}}{2cdotsqrt{pi}cdottext{n}}right)right}tag4$$




        Now, your task is to simplify and to take:



        $$lim_{text{m}toinfty}mathscr{I}_{spacetext{n}}left(text{m}space,spacetext{s}right)tag5$$





        Try to prove:



        $$lim_{text{m}toinfty}text{erf}left(frac{2cdotpicdottext{n}^2cdottext{m}+text{s}}{2cdotsqrt{pi}cdottext{n}}right)=1tag6$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 18 '17 at 15:55









        JanJan

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            張江高科駅

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            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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