Solving a double integral/Finding a normal distribution












0












$begingroup$


Let $sigma^2,alpha,s,t>0$, $Binmathcal{B}(mathbb{R})$ and $xinmathbb{R}$. Consider following integral



begin{align}frac{1}{sqrt{2pifrac{sigma^2}{2alpha}(1-e^{-alpha t})}sqrt{2pifrac{sigma^2}{2alpha}(1-e^{-alpha s})}}cdot\ int_mathbb{R}bigg[int_BexpBig(&-frac{(z-ye^{-alpha t})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha t})}Big)dzbigg]quadexpBig(-frac{(y-xe^{-alpha s})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha s})}Big)dyend{align}



I want to find it being equal to



$$frac{1}{sqrt{2pifrac{sigma^2}{2alpha}(1-e^{-2alpha(t+s)})}}int_B expBig(-frac{(z-xe^{-alpha (t+s)})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha(t+s)})}Big)dz$$



which is a normal distribution. I think one can integrate over $B-ye^{-alpha t}$ and then integrate by substitution, but I struggle finding the result. Thank you for any help!






real-analysis calculus probability analysis probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you change the order of integration and focus on the inner integral (with respect to $y$) you will see that the integrand is $c exp(-(text{quadratic function of $y$}))$, where there will be other terms involving $z$ and constants like $alpha$, $sigma^2$, $t$, $s$. By completing the square in the exponent, you can obtain another Gaussian density that allows you to integrate with respect to $y$ over $mathbb{R}$. What remains should be the last line.
    $endgroup$
    – angryavian
    Dec 20 '18 at 20:24










  • $begingroup$
    I do not really understand. Do you mean I should put the two exponential functions together and then integrate over $B$ with respect to $dy$ ?
    $endgroup$
    – user628255
    Dec 20 '18 at 20:41












  • $begingroup$
    Yes, but integrate over $mathbb{R}$ with respect to $dy$.
    $endgroup$
    – angryavian
    Dec 20 '18 at 21:09










  • $begingroup$
    I do not get any solution... Can you give it a try? And maybe post a picture?
    $endgroup$
    – user628255
    Dec 20 '18 at 21:45
















0












$begingroup$


Let $sigma^2,alpha,s,t>0$, $Binmathcal{B}(mathbb{R})$ and $xinmathbb{R}$. Consider following integral



begin{align}frac{1}{sqrt{2pifrac{sigma^2}{2alpha}(1-e^{-alpha t})}sqrt{2pifrac{sigma^2}{2alpha}(1-e^{-alpha s})}}cdot\ int_mathbb{R}bigg[int_BexpBig(&-frac{(z-ye^{-alpha t})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha t})}Big)dzbigg]quadexpBig(-frac{(y-xe^{-alpha s})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha s})}Big)dyend{align}



I want to find it being equal to



$$frac{1}{sqrt{2pifrac{sigma^2}{2alpha}(1-e^{-2alpha(t+s)})}}int_B expBig(-frac{(z-xe^{-alpha (t+s)})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha(t+s)})}Big)dz$$



which is a normal distribution. I think one can integrate over $B-ye^{-alpha t}$ and then integrate by substitution, but I struggle finding the result. Thank you for any help!






real-analysis calculus probability analysis probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you change the order of integration and focus on the inner integral (with respect to $y$) you will see that the integrand is $c exp(-(text{quadratic function of $y$}))$, where there will be other terms involving $z$ and constants like $alpha$, $sigma^2$, $t$, $s$. By completing the square in the exponent, you can obtain another Gaussian density that allows you to integrate with respect to $y$ over $mathbb{R}$. What remains should be the last line.
    $endgroup$
    – angryavian
    Dec 20 '18 at 20:24










  • $begingroup$
    I do not really understand. Do you mean I should put the two exponential functions together and then integrate over $B$ with respect to $dy$ ?
    $endgroup$
    – user628255
    Dec 20 '18 at 20:41












  • $begingroup$
    Yes, but integrate over $mathbb{R}$ with respect to $dy$.
    $endgroup$
    – angryavian
    Dec 20 '18 at 21:09










  • $begingroup$
    I do not get any solution... Can you give it a try? And maybe post a picture?
    $endgroup$
    – user628255
    Dec 20 '18 at 21:45














0












0








0





$begingroup$


Let $sigma^2,alpha,s,t>0$, $Binmathcal{B}(mathbb{R})$ and $xinmathbb{R}$. Consider following integral



begin{align}frac{1}{sqrt{2pifrac{sigma^2}{2alpha}(1-e^{-alpha t})}sqrt{2pifrac{sigma^2}{2alpha}(1-e^{-alpha s})}}cdot\ int_mathbb{R}bigg[int_BexpBig(&-frac{(z-ye^{-alpha t})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha t})}Big)dzbigg]quadexpBig(-frac{(y-xe^{-alpha s})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha s})}Big)dyend{align}



I want to find it being equal to



$$frac{1}{sqrt{2pifrac{sigma^2}{2alpha}(1-e^{-2alpha(t+s)})}}int_B expBig(-frac{(z-xe^{-alpha (t+s)})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha(t+s)})}Big)dz$$



which is a normal distribution. I think one can integrate over $B-ye^{-alpha t}$ and then integrate by substitution, but I struggle finding the result. Thank you for any help!






real-analysis calculus probability analysis probability-distributions







share|cite|improve this question











$endgroup$




Let $sigma^2,alpha,s,t>0$, $Binmathcal{B}(mathbb{R})$ and $xinmathbb{R}$. Consider following integral



begin{align}frac{1}{sqrt{2pifrac{sigma^2}{2alpha}(1-e^{-alpha t})}sqrt{2pifrac{sigma^2}{2alpha}(1-e^{-alpha s})}}cdot\ int_mathbb{R}bigg[int_BexpBig(&-frac{(z-ye^{-alpha t})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha t})}Big)dzbigg]quadexpBig(-frac{(y-xe^{-alpha s})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha s})}Big)dyend{align}



I want to find it being equal to



$$frac{1}{sqrt{2pifrac{sigma^2}{2alpha}(1-e^{-2alpha(t+s)})}}int_B expBig(-frac{(z-xe^{-alpha (t+s)})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha(t+s)})}Big)dz$$



which is a normal distribution. I think one can integrate over $B-ye^{-alpha t}$ and then integrate by substitution, but I struggle finding the result. Thank you for any help!






real-analysis calculus probability analysis probability-distributions




real-analysis calculus probability analysis probability-distributions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 20:39







user628255

















asked Dec 20 '18 at 20:16









user628255user628255

224




224












  • $begingroup$
    If you change the order of integration and focus on the inner integral (with respect to $y$) you will see that the integrand is $c exp(-(text{quadratic function of $y$}))$, where there will be other terms involving $z$ and constants like $alpha$, $sigma^2$, $t$, $s$. By completing the square in the exponent, you can obtain another Gaussian density that allows you to integrate with respect to $y$ over $mathbb{R}$. What remains should be the last line.
    $endgroup$
    – angryavian
    Dec 20 '18 at 20:24










  • $begingroup$
    I do not really understand. Do you mean I should put the two exponential functions together and then integrate over $B$ with respect to $dy$ ?
    $endgroup$
    – user628255
    Dec 20 '18 at 20:41












  • $begingroup$
    Yes, but integrate over $mathbb{R}$ with respect to $dy$.
    $endgroup$
    – angryavian
    Dec 20 '18 at 21:09










  • $begingroup$
    I do not get any solution... Can you give it a try? And maybe post a picture?
    $endgroup$
    – user628255
    Dec 20 '18 at 21:45


















  • $begingroup$
    If you change the order of integration and focus on the inner integral (with respect to $y$) you will see that the integrand is $c exp(-(text{quadratic function of $y$}))$, where there will be other terms involving $z$ and constants like $alpha$, $sigma^2$, $t$, $s$. By completing the square in the exponent, you can obtain another Gaussian density that allows you to integrate with respect to $y$ over $mathbb{R}$. What remains should be the last line.
    $endgroup$
    – angryavian
    Dec 20 '18 at 20:24










  • $begingroup$
    I do not really understand. Do you mean I should put the two exponential functions together and then integrate over $B$ with respect to $dy$ ?
    $endgroup$
    – user628255
    Dec 20 '18 at 20:41












  • $begingroup$
    Yes, but integrate over $mathbb{R}$ with respect to $dy$.
    $endgroup$
    – angryavian
    Dec 20 '18 at 21:09










  • $begingroup$
    I do not get any solution... Can you give it a try? And maybe post a picture?
    $endgroup$
    – user628255
    Dec 20 '18 at 21:45
















$begingroup$
If you change the order of integration and focus on the inner integral (with respect to $y$) you will see that the integrand is $c exp(-(text{quadratic function of $y$}))$, where there will be other terms involving $z$ and constants like $alpha$, $sigma^2$, $t$, $s$. By completing the square in the exponent, you can obtain another Gaussian density that allows you to integrate with respect to $y$ over $mathbb{R}$. What remains should be the last line.
$endgroup$
– angryavian
Dec 20 '18 at 20:24




$begingroup$
If you change the order of integration and focus on the inner integral (with respect to $y$) you will see that the integrand is $c exp(-(text{quadratic function of $y$}))$, where there will be other terms involving $z$ and constants like $alpha$, $sigma^2$, $t$, $s$. By completing the square in the exponent, you can obtain another Gaussian density that allows you to integrate with respect to $y$ over $mathbb{R}$. What remains should be the last line.
$endgroup$
– angryavian
Dec 20 '18 at 20:24












$begingroup$
I do not really understand. Do you mean I should put the two exponential functions together and then integrate over $B$ with respect to $dy$ ?
$endgroup$
– user628255
Dec 20 '18 at 20:41






$begingroup$
I do not really understand. Do you mean I should put the two exponential functions together and then integrate over $B$ with respect to $dy$ ?
$endgroup$
– user628255
Dec 20 '18 at 20:41














$begingroup$
Yes, but integrate over $mathbb{R}$ with respect to $dy$.
$endgroup$
– angryavian
Dec 20 '18 at 21:09




$begingroup$
Yes, but integrate over $mathbb{R}$ with respect to $dy$.
$endgroup$
– angryavian
Dec 20 '18 at 21:09












$begingroup$
I do not get any solution... Can you give it a try? And maybe post a picture?
$endgroup$
– user628255
Dec 20 '18 at 21:45




$begingroup$
I do not get any solution... Can you give it a try? And maybe post a picture?
$endgroup$
– user628255
Dec 20 '18 at 21:45










1 Answer
1






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oldest

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1












$begingroup$

Here is an approach that avoids directly manipulating the integrals (which is quite painful).



Let $$Y sim N(xe^{-alpha s}, frac{sigma^2}{2alpha}(1-e^{-2alpha s}).$$
Let the conditional distribution of $Z$ given $Y=y$ be $$N(ye^{-alpha t}, frac{sigma^2}{2alpha} (1-e^{-2 alpha t})).$$
Note that the joint distribution satisfies $f_{Y,Z}(y,z) = f_{Z mid Y=y}(z) cdot f_Y(y)$. This is precisely your integrand. In fact, the double integral is computing $P(Z in B)$.



To compute this, one could instead find the marginal distribution of $Z$. I believe one can show that $(Y,Z)$ is bivariate Gaussian, so $Z$ is also Gaussian. Its expectation is $$E[Z] = E[E[Z mid Y]] = E[Ye^{-alpha t}] = xe^{-alpha(s+t)}$$
and its variance is $$text{Var}(Z) = E[text{Var}(Z mid Y)] + text{Var}(E[Z mid Y]) = frac{sigma^2}{2alpha}(1-e^{-2alpha t}) + e^{-2alpha t} frac{sigma^2}{2alpha}(1-e^{-2alpha s}) = frac{sigma^2}{2alpha}(1-e^{-2alpha (s+t)}).$$
Thus you can write down the density $f_Z$ of $Z$, and have $P(Z in B) = int_B f_Z(z) , dz$. This is the goal integral.






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    1 Answer
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    active

    oldest

    votes









    1












    $begingroup$

    Here is an approach that avoids directly manipulating the integrals (which is quite painful).



    Let $$Y sim N(xe^{-alpha s}, frac{sigma^2}{2alpha}(1-e^{-2alpha s}).$$
    Let the conditional distribution of $Z$ given $Y=y$ be $$N(ye^{-alpha t}, frac{sigma^2}{2alpha} (1-e^{-2 alpha t})).$$
    Note that the joint distribution satisfies $f_{Y,Z}(y,z) = f_{Z mid Y=y}(z) cdot f_Y(y)$. This is precisely your integrand. In fact, the double integral is computing $P(Z in B)$.



    To compute this, one could instead find the marginal distribution of $Z$. I believe one can show that $(Y,Z)$ is bivariate Gaussian, so $Z$ is also Gaussian. Its expectation is $$E[Z] = E[E[Z mid Y]] = E[Ye^{-alpha t}] = xe^{-alpha(s+t)}$$
    and its variance is $$text{Var}(Z) = E[text{Var}(Z mid Y)] + text{Var}(E[Z mid Y]) = frac{sigma^2}{2alpha}(1-e^{-2alpha t}) + e^{-2alpha t} frac{sigma^2}{2alpha}(1-e^{-2alpha s}) = frac{sigma^2}{2alpha}(1-e^{-2alpha (s+t)}).$$
    Thus you can write down the density $f_Z$ of $Z$, and have $P(Z in B) = int_B f_Z(z) , dz$. This is the goal integral.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Here is an approach that avoids directly manipulating the integrals (which is quite painful).



      Let $$Y sim N(xe^{-alpha s}, frac{sigma^2}{2alpha}(1-e^{-2alpha s}).$$
      Let the conditional distribution of $Z$ given $Y=y$ be $$N(ye^{-alpha t}, frac{sigma^2}{2alpha} (1-e^{-2 alpha t})).$$
      Note that the joint distribution satisfies $f_{Y,Z}(y,z) = f_{Z mid Y=y}(z) cdot f_Y(y)$. This is precisely your integrand. In fact, the double integral is computing $P(Z in B)$.



      To compute this, one could instead find the marginal distribution of $Z$. I believe one can show that $(Y,Z)$ is bivariate Gaussian, so $Z$ is also Gaussian. Its expectation is $$E[Z] = E[E[Z mid Y]] = E[Ye^{-alpha t}] = xe^{-alpha(s+t)}$$
      and its variance is $$text{Var}(Z) = E[text{Var}(Z mid Y)] + text{Var}(E[Z mid Y]) = frac{sigma^2}{2alpha}(1-e^{-2alpha t}) + e^{-2alpha t} frac{sigma^2}{2alpha}(1-e^{-2alpha s}) = frac{sigma^2}{2alpha}(1-e^{-2alpha (s+t)}).$$
      Thus you can write down the density $f_Z$ of $Z$, and have $P(Z in B) = int_B f_Z(z) , dz$. This is the goal integral.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Here is an approach that avoids directly manipulating the integrals (which is quite painful).



        Let $$Y sim N(xe^{-alpha s}, frac{sigma^2}{2alpha}(1-e^{-2alpha s}).$$
        Let the conditional distribution of $Z$ given $Y=y$ be $$N(ye^{-alpha t}, frac{sigma^2}{2alpha} (1-e^{-2 alpha t})).$$
        Note that the joint distribution satisfies $f_{Y,Z}(y,z) = f_{Z mid Y=y}(z) cdot f_Y(y)$. This is precisely your integrand. In fact, the double integral is computing $P(Z in B)$.



        To compute this, one could instead find the marginal distribution of $Z$. I believe one can show that $(Y,Z)$ is bivariate Gaussian, so $Z$ is also Gaussian. Its expectation is $$E[Z] = E[E[Z mid Y]] = E[Ye^{-alpha t}] = xe^{-alpha(s+t)}$$
        and its variance is $$text{Var}(Z) = E[text{Var}(Z mid Y)] + text{Var}(E[Z mid Y]) = frac{sigma^2}{2alpha}(1-e^{-2alpha t}) + e^{-2alpha t} frac{sigma^2}{2alpha}(1-e^{-2alpha s}) = frac{sigma^2}{2alpha}(1-e^{-2alpha (s+t)}).$$
        Thus you can write down the density $f_Z$ of $Z$, and have $P(Z in B) = int_B f_Z(z) , dz$. This is the goal integral.






        share|cite|improve this answer









        $endgroup$



        Here is an approach that avoids directly manipulating the integrals (which is quite painful).



        Let $$Y sim N(xe^{-alpha s}, frac{sigma^2}{2alpha}(1-e^{-2alpha s}).$$
        Let the conditional distribution of $Z$ given $Y=y$ be $$N(ye^{-alpha t}, frac{sigma^2}{2alpha} (1-e^{-2 alpha t})).$$
        Note that the joint distribution satisfies $f_{Y,Z}(y,z) = f_{Z mid Y=y}(z) cdot f_Y(y)$. This is precisely your integrand. In fact, the double integral is computing $P(Z in B)$.



        To compute this, one could instead find the marginal distribution of $Z$. I believe one can show that $(Y,Z)$ is bivariate Gaussian, so $Z$ is also Gaussian. Its expectation is $$E[Z] = E[E[Z mid Y]] = E[Ye^{-alpha t}] = xe^{-alpha(s+t)}$$
        and its variance is $$text{Var}(Z) = E[text{Var}(Z mid Y)] + text{Var}(E[Z mid Y]) = frac{sigma^2}{2alpha}(1-e^{-2alpha t}) + e^{-2alpha t} frac{sigma^2}{2alpha}(1-e^{-2alpha s}) = frac{sigma^2}{2alpha}(1-e^{-2alpha (s+t)}).$$
        Thus you can write down the density $f_Z$ of $Z$, and have $P(Z in B) = int_B f_Z(z) , dz$. This is the goal integral.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 20 '18 at 22:31









        angryavianangryavian

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