Compute $mathbb{E}big[exp(XY+X)big]$ where $X, Y$ are independent uniformly distributed over $[0,1]$ r.v's.












3












$begingroup$



Compute $mathbb{E}big[exp(XY+X)big]$ where $X, Y$ are independent uniformly distributed over $[0,1]$ r.v's.




I first computed $mathbb{E}big[exp(XY+X)mid Xbig]$.



Because $X$ and $Y$ are independent then $$mathbb{E}big[exp(XY+X)mid Xbig] = phi(X),,$$
where $$phi(x) =mathbb{E}big[exp(xY+x)big] = int_0^1 exp(xy+x)dy = frac{exp x(exp x - 1)}{x}.$$
so $$mathbb{E}big[exp(XY+X)big] = int_0^1 frac{exp x(exp x - 1)}{x} dx,,$$
which I don't know how to compute.



Could someone check if I'm good till now and maybe help me continue? Thanks!






integration probability-theory random-variables conditional-expectation expected-value







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You are good :-) no mistake.
    $endgroup$
    – Math-fun
    Jan 8 at 20:05










  • $begingroup$
    You received two answers with different numerical results. Have you checked out the discrepancy?
    $endgroup$
    – herb steinberg
    Jan 10 at 18:28






  • 1




    $begingroup$
    @rapidracim I have corrected my answer and there is no discrepancy between the two results.
    $endgroup$
    – herb steinberg
    Jan 14 at 20:46










  • $begingroup$
    @herbsteinberg I checked your edit, I get it now.
    $endgroup$
    – rapidracim
    Jan 14 at 21:39
















3












$begingroup$



Compute $mathbb{E}big[exp(XY+X)big]$ where $X, Y$ are independent uniformly distributed over $[0,1]$ r.v's.




I first computed $mathbb{E}big[exp(XY+X)mid Xbig]$.



Because $X$ and $Y$ are independent then $$mathbb{E}big[exp(XY+X)mid Xbig] = phi(X),,$$
where $$phi(x) =mathbb{E}big[exp(xY+x)big] = int_0^1 exp(xy+x)dy = frac{exp x(exp x - 1)}{x}.$$
so $$mathbb{E}big[exp(XY+X)big] = int_0^1 frac{exp x(exp x - 1)}{x} dx,,$$
which I don't know how to compute.



Could someone check if I'm good till now and maybe help me continue? Thanks!






integration probability-theory random-variables conditional-expectation expected-value







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You are good :-) no mistake.
    $endgroup$
    – Math-fun
    Jan 8 at 20:05










  • $begingroup$
    You received two answers with different numerical results. Have you checked out the discrepancy?
    $endgroup$
    – herb steinberg
    Jan 10 at 18:28






  • 1




    $begingroup$
    @rapidracim I have corrected my answer and there is no discrepancy between the two results.
    $endgroup$
    – herb steinberg
    Jan 14 at 20:46










  • $begingroup$
    @herbsteinberg I checked your edit, I get it now.
    $endgroup$
    – rapidracim
    Jan 14 at 21:39














3












3








3





$begingroup$



Compute $mathbb{E}big[exp(XY+X)big]$ where $X, Y$ are independent uniformly distributed over $[0,1]$ r.v's.




I first computed $mathbb{E}big[exp(XY+X)mid Xbig]$.



Because $X$ and $Y$ are independent then $$mathbb{E}big[exp(XY+X)mid Xbig] = phi(X),,$$
where $$phi(x) =mathbb{E}big[exp(xY+x)big] = int_0^1 exp(xy+x)dy = frac{exp x(exp x - 1)}{x}.$$
so $$mathbb{E}big[exp(XY+X)big] = int_0^1 frac{exp x(exp x - 1)}{x} dx,,$$
which I don't know how to compute.



Could someone check if I'm good till now and maybe help me continue? Thanks!






integration probability-theory random-variables conditional-expectation expected-value







share|cite|improve this question











$endgroup$





Compute $mathbb{E}big[exp(XY+X)big]$ where $X, Y$ are independent uniformly distributed over $[0,1]$ r.v's.




I first computed $mathbb{E}big[exp(XY+X)mid Xbig]$.



Because $X$ and $Y$ are independent then $$mathbb{E}big[exp(XY+X)mid Xbig] = phi(X),,$$
where $$phi(x) =mathbb{E}big[exp(xY+x)big] = int_0^1 exp(xy+x)dy = frac{exp x(exp x - 1)}{x}.$$
so $$mathbb{E}big[exp(XY+X)big] = int_0^1 frac{exp x(exp x - 1)}{x} dx,,$$
which I don't know how to compute.



Could someone check if I'm good till now and maybe help me continue? Thanks!






integration probability-theory random-variables conditional-expectation expected-value




integration probability-theory random-variables conditional-expectation expected-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 22:06









Batominovski

1




1










asked Jan 8 at 19:57









rapidracimrapidracim

1,6141319




1,6141319








  • 1




    $begingroup$
    You are good :-) no mistake.
    $endgroup$
    – Math-fun
    Jan 8 at 20:05










  • $begingroup$
    You received two answers with different numerical results. Have you checked out the discrepancy?
    $endgroup$
    – herb steinberg
    Jan 10 at 18:28






  • 1




    $begingroup$
    @rapidracim I have corrected my answer and there is no discrepancy between the two results.
    $endgroup$
    – herb steinberg
    Jan 14 at 20:46










  • $begingroup$
    @herbsteinberg I checked your edit, I get it now.
    $endgroup$
    – rapidracim
    Jan 14 at 21:39














  • 1




    $begingroup$
    You are good :-) no mistake.
    $endgroup$
    – Math-fun
    Jan 8 at 20:05










  • $begingroup$
    You received two answers with different numerical results. Have you checked out the discrepancy?
    $endgroup$
    – herb steinberg
    Jan 10 at 18:28






  • 1




    $begingroup$
    @rapidracim I have corrected my answer and there is no discrepancy between the two results.
    $endgroup$
    – herb steinberg
    Jan 14 at 20:46










  • $begingroup$
    @herbsteinberg I checked your edit, I get it now.
    $endgroup$
    – rapidracim
    Jan 14 at 21:39








1




1




$begingroup$
You are good :-) no mistake.
$endgroup$
– Math-fun
Jan 8 at 20:05




$begingroup$
You are good :-) no mistake.
$endgroup$
– Math-fun
Jan 8 at 20:05












$begingroup$
You received two answers with different numerical results. Have you checked out the discrepancy?
$endgroup$
– herb steinberg
Jan 10 at 18:28




$begingroup$
You received two answers with different numerical results. Have you checked out the discrepancy?
$endgroup$
– herb steinberg
Jan 10 at 18:28




1




1




$begingroup$
@rapidracim I have corrected my answer and there is no discrepancy between the two results.
$endgroup$
– herb steinberg
Jan 14 at 20:46




$begingroup$
@rapidracim I have corrected my answer and there is no discrepancy between the two results.
$endgroup$
– herb steinberg
Jan 14 at 20:46












$begingroup$
@herbsteinberg I checked your edit, I get it now.
$endgroup$
– rapidracim
Jan 14 at 21:39




$begingroup$
@herbsteinberg I checked your edit, I get it now.
$endgroup$
– rapidracim
Jan 14 at 21:39










2 Answers
2






active

oldest

votes


















2












$begingroup$

You have done great so far. But I dont think there is an elementary expression for the last integral. But you can note that the exponential integral $operatorname{Ei}$ is given by
$$operatorname{Ei}(t)=mathrel{-!!!!!!;!!int}_{-infty}^{t}frac{e^{s}}{s}ds.$$
(Here $mathrel{-!!!!!;!!int}$ is the Cauchy principal value integral.)
In other words,
$$intfrac{e^{s}}{s}ds=operatorname{Ei}(s)+C.$$
This also shows that
$$intfrac{e^{ks}}{s}ds=operatorname{Ei}(ks)+C.$$



The required integral is
begin{align}I&=int_0^1frac{e^x(e^x-1)}{x}dx=lim_{epsilonsearrow0}int_{epsilon}^1left(frac{e^{2x}}{x}dx-int_epsilon^1frac{e^x}{x}dxright)\
&=lim_{epsilonsearrow0}Big(big(operatorname{Ei}(2)-operatorname{Ei}(2epsilon)big)-big(operatorname{Ei}(1)-operatorname{Ei}(epsilon)big)Big)\
&=operatorname{Ei}(2)-operatorname{Ei}(1)-lim_{epsilonsearrow0}big(operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)big).end{align}
Now
$$operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)=int_{epsilon}^{2epsilon}frac{e^s}{s}ds=int_epsilon^{2epsilon}frac{1+O(epsilon)}{s}ds=int_{epsilon}^{2epsilon}frac{ds}{s}+O(epsilon)=ln 2+O(epsilon).$$
This gives
$$I=operatorname{Ei}(2)-operatorname{Ei}(1)-ln2approx 2.366.$$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    You can't get an explicit answer. However let $u=exp(x)$ then $du=exp(x)dx$ and your integral becomes $int_1^e frac{u-1}{ln(u)}du=text{li}(e^2)-text{li}(e)=int_e^{e^2}frac{1}{ln(u)}du$. Unfortunately there is no explicit formula for $text{li}(u)$.
    I calculated this result and got I=3.0591165.

    In addition, due to the singularity of the integrand at $u=1$, it is necessary to subtract $lim_{epsilonto 0}int_{(1+epsilon)}^{(1+epsilon)^2}frac{1}{ln(u)}du=ln(2)=0.693147181$

    Net result$=2.365969319$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @ Batominovski What was the edit? I didn't see any changes!
      $endgroup$
      – herb steinberg
      Jan 8 at 22:34











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    2 Answers
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    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    You have done great so far. But I dont think there is an elementary expression for the last integral. But you can note that the exponential integral $operatorname{Ei}$ is given by
    $$operatorname{Ei}(t)=mathrel{-!!!!!!;!!int}_{-infty}^{t}frac{e^{s}}{s}ds.$$
    (Here $mathrel{-!!!!!;!!int}$ is the Cauchy principal value integral.)
    In other words,
    $$intfrac{e^{s}}{s}ds=operatorname{Ei}(s)+C.$$
    This also shows that
    $$intfrac{e^{ks}}{s}ds=operatorname{Ei}(ks)+C.$$



    The required integral is
    begin{align}I&=int_0^1frac{e^x(e^x-1)}{x}dx=lim_{epsilonsearrow0}int_{epsilon}^1left(frac{e^{2x}}{x}dx-int_epsilon^1frac{e^x}{x}dxright)\
    &=lim_{epsilonsearrow0}Big(big(operatorname{Ei}(2)-operatorname{Ei}(2epsilon)big)-big(operatorname{Ei}(1)-operatorname{Ei}(epsilon)big)Big)\
    &=operatorname{Ei}(2)-operatorname{Ei}(1)-lim_{epsilonsearrow0}big(operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)big).end{align}
    Now
    $$operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)=int_{epsilon}^{2epsilon}frac{e^s}{s}ds=int_epsilon^{2epsilon}frac{1+O(epsilon)}{s}ds=int_{epsilon}^{2epsilon}frac{ds}{s}+O(epsilon)=ln 2+O(epsilon).$$
    This gives
    $$I=operatorname{Ei}(2)-operatorname{Ei}(1)-ln2approx 2.366.$$






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      You have done great so far. But I dont think there is an elementary expression for the last integral. But you can note that the exponential integral $operatorname{Ei}$ is given by
      $$operatorname{Ei}(t)=mathrel{-!!!!!!;!!int}_{-infty}^{t}frac{e^{s}}{s}ds.$$
      (Here $mathrel{-!!!!!;!!int}$ is the Cauchy principal value integral.)
      In other words,
      $$intfrac{e^{s}}{s}ds=operatorname{Ei}(s)+C.$$
      This also shows that
      $$intfrac{e^{ks}}{s}ds=operatorname{Ei}(ks)+C.$$



      The required integral is
      begin{align}I&=int_0^1frac{e^x(e^x-1)}{x}dx=lim_{epsilonsearrow0}int_{epsilon}^1left(frac{e^{2x}}{x}dx-int_epsilon^1frac{e^x}{x}dxright)\
      &=lim_{epsilonsearrow0}Big(big(operatorname{Ei}(2)-operatorname{Ei}(2epsilon)big)-big(operatorname{Ei}(1)-operatorname{Ei}(epsilon)big)Big)\
      &=operatorname{Ei}(2)-operatorname{Ei}(1)-lim_{epsilonsearrow0}big(operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)big).end{align}
      Now
      $$operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)=int_{epsilon}^{2epsilon}frac{e^s}{s}ds=int_epsilon^{2epsilon}frac{1+O(epsilon)}{s}ds=int_{epsilon}^{2epsilon}frac{ds}{s}+O(epsilon)=ln 2+O(epsilon).$$
      This gives
      $$I=operatorname{Ei}(2)-operatorname{Ei}(1)-ln2approx 2.366.$$






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        You have done great so far. But I dont think there is an elementary expression for the last integral. But you can note that the exponential integral $operatorname{Ei}$ is given by
        $$operatorname{Ei}(t)=mathrel{-!!!!!!;!!int}_{-infty}^{t}frac{e^{s}}{s}ds.$$
        (Here $mathrel{-!!!!!;!!int}$ is the Cauchy principal value integral.)
        In other words,
        $$intfrac{e^{s}}{s}ds=operatorname{Ei}(s)+C.$$
        This also shows that
        $$intfrac{e^{ks}}{s}ds=operatorname{Ei}(ks)+C.$$



        The required integral is
        begin{align}I&=int_0^1frac{e^x(e^x-1)}{x}dx=lim_{epsilonsearrow0}int_{epsilon}^1left(frac{e^{2x}}{x}dx-int_epsilon^1frac{e^x}{x}dxright)\
        &=lim_{epsilonsearrow0}Big(big(operatorname{Ei}(2)-operatorname{Ei}(2epsilon)big)-big(operatorname{Ei}(1)-operatorname{Ei}(epsilon)big)Big)\
        &=operatorname{Ei}(2)-operatorname{Ei}(1)-lim_{epsilonsearrow0}big(operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)big).end{align}
        Now
        $$operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)=int_{epsilon}^{2epsilon}frac{e^s}{s}ds=int_epsilon^{2epsilon}frac{1+O(epsilon)}{s}ds=int_{epsilon}^{2epsilon}frac{ds}{s}+O(epsilon)=ln 2+O(epsilon).$$
        This gives
        $$I=operatorname{Ei}(2)-operatorname{Ei}(1)-ln2approx 2.366.$$






        share|cite|improve this answer











        $endgroup$



        You have done great so far. But I dont think there is an elementary expression for the last integral. But you can note that the exponential integral $operatorname{Ei}$ is given by
        $$operatorname{Ei}(t)=mathrel{-!!!!!!;!!int}_{-infty}^{t}frac{e^{s}}{s}ds.$$
        (Here $mathrel{-!!!!!;!!int}$ is the Cauchy principal value integral.)
        In other words,
        $$intfrac{e^{s}}{s}ds=operatorname{Ei}(s)+C.$$
        This also shows that
        $$intfrac{e^{ks}}{s}ds=operatorname{Ei}(ks)+C.$$



        The required integral is
        begin{align}I&=int_0^1frac{e^x(e^x-1)}{x}dx=lim_{epsilonsearrow0}int_{epsilon}^1left(frac{e^{2x}}{x}dx-int_epsilon^1frac{e^x}{x}dxright)\
        &=lim_{epsilonsearrow0}Big(big(operatorname{Ei}(2)-operatorname{Ei}(2epsilon)big)-big(operatorname{Ei}(1)-operatorname{Ei}(epsilon)big)Big)\
        &=operatorname{Ei}(2)-operatorname{Ei}(1)-lim_{epsilonsearrow0}big(operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)big).end{align}
        Now
        $$operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)=int_{epsilon}^{2epsilon}frac{e^s}{s}ds=int_epsilon^{2epsilon}frac{1+O(epsilon)}{s}ds=int_{epsilon}^{2epsilon}frac{ds}{s}+O(epsilon)=ln 2+O(epsilon).$$
        This gives
        $$I=operatorname{Ei}(2)-operatorname{Ei}(1)-ln2approx 2.366.$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 8 at 20:58

























        answered Jan 8 at 20:50







        user614671






























            0












            $begingroup$

            You can't get an explicit answer. However let $u=exp(x)$ then $du=exp(x)dx$ and your integral becomes $int_1^e frac{u-1}{ln(u)}du=text{li}(e^2)-text{li}(e)=int_e^{e^2}frac{1}{ln(u)}du$. Unfortunately there is no explicit formula for $text{li}(u)$.
            I calculated this result and got I=3.0591165.

            In addition, due to the singularity of the integrand at $u=1$, it is necessary to subtract $lim_{epsilonto 0}int_{(1+epsilon)}^{(1+epsilon)^2}frac{1}{ln(u)}du=ln(2)=0.693147181$

            Net result$=2.365969319$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @ Batominovski What was the edit? I didn't see any changes!
              $endgroup$
              – herb steinberg
              Jan 8 at 22:34
















            0












            $begingroup$

            You can't get an explicit answer. However let $u=exp(x)$ then $du=exp(x)dx$ and your integral becomes $int_1^e frac{u-1}{ln(u)}du=text{li}(e^2)-text{li}(e)=int_e^{e^2}frac{1}{ln(u)}du$. Unfortunately there is no explicit formula for $text{li}(u)$.
            I calculated this result and got I=3.0591165.

            In addition, due to the singularity of the integrand at $u=1$, it is necessary to subtract $lim_{epsilonto 0}int_{(1+epsilon)}^{(1+epsilon)^2}frac{1}{ln(u)}du=ln(2)=0.693147181$

            Net result$=2.365969319$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @ Batominovski What was the edit? I didn't see any changes!
              $endgroup$
              – herb steinberg
              Jan 8 at 22:34














            0












            0








            0





            $begingroup$

            You can't get an explicit answer. However let $u=exp(x)$ then $du=exp(x)dx$ and your integral becomes $int_1^e frac{u-1}{ln(u)}du=text{li}(e^2)-text{li}(e)=int_e^{e^2}frac{1}{ln(u)}du$. Unfortunately there is no explicit formula for $text{li}(u)$.
            I calculated this result and got I=3.0591165.

            In addition, due to the singularity of the integrand at $u=1$, it is necessary to subtract $lim_{epsilonto 0}int_{(1+epsilon)}^{(1+epsilon)^2}frac{1}{ln(u)}du=ln(2)=0.693147181$

            Net result$=2.365969319$.






            share|cite|improve this answer











            $endgroup$



            You can't get an explicit answer. However let $u=exp(x)$ then $du=exp(x)dx$ and your integral becomes $int_1^e frac{u-1}{ln(u)}du=text{li}(e^2)-text{li}(e)=int_e^{e^2}frac{1}{ln(u)}du$. Unfortunately there is no explicit formula for $text{li}(u)$.
            I calculated this result and got I=3.0591165.

            In addition, due to the singularity of the integrand at $u=1$, it is necessary to subtract $lim_{epsilonto 0}int_{(1+epsilon)}^{(1+epsilon)^2}frac{1}{ln(u)}du=ln(2)=0.693147181$

            Net result$=2.365969319$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 14 at 20:44

























            answered Jan 8 at 21:06









            herb steinbergherb steinberg

            2,5932310




            2,5932310












            • $begingroup$
              @ Batominovski What was the edit? I didn't see any changes!
              $endgroup$
              – herb steinberg
              Jan 8 at 22:34


















            • $begingroup$
              @ Batominovski What was the edit? I didn't see any changes!
              $endgroup$
              – herb steinberg
              Jan 8 at 22:34
















            $begingroup$
            @ Batominovski What was the edit? I didn't see any changes!
            $endgroup$
            – herb steinberg
            Jan 8 at 22:34




            $begingroup$
            @ Batominovski What was the edit? I didn't see any changes!
            $endgroup$
            – herb steinberg
            Jan 8 at 22:34


















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            A Topological Invariant for $pi_3(U(n))$

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            3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
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