Integral involving the normal distribution and its CDF times x^2
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I'm trying to evaluate the integral
begin{equation}
int_{-infty}^{infty} s^2 f_{s_1} (s) Phi Big(dfrac{s}{sigma_n} Big) , ds.
end{equation}
where begin{equation}
f_{bar{s} _l}(s) = dfrac{1}{sqrt{2pi sigma_{1}^2}} e^{tfrac{text{-}(s-{s}_l)^2}{2sigma_1^2}},
end{equation}
and $Phi(x)$ is the CDF function, any ideas?
integration normal-distribution
asked Nov 22 at 5:11
Diana
112
add a comment |
up vote
0
down vote
favorite
I'm trying to evaluate the integral
begin{equation}
int_{-infty}^{infty} s^2 f_{s_1} (s) Phi Big(dfrac{s}{sigma_n} Big) , ds.
end{equation}
where begin{equation}
f_{bar{s} _l}(s) = dfrac{1}{sqrt{2pi sigma_{1}^2}} e^{tfrac{text{-}(s-{s}_l)^2}{2sigma_1^2}},
end{equation}
and $Phi(x)$ is the CDF function, any ideas?
integration normal-distribution
asked Nov 22 at 5:11
Diana
112
Not indefinite, but improper integral.
– StubbornAtom
Nov 22 at 5:37
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to evaluate the integral
begin{equation}
int_{-infty}^{infty} s^2 f_{s_1} (s) Phi Big(dfrac{s}{sigma_n} Big) , ds.
end{equation}
where begin{equation}
f_{bar{s} _l}(s) = dfrac{1}{sqrt{2pi sigma_{1}^2}} e^{tfrac{text{-}(s-{s}_l)^2}{2sigma_1^2}},
end{equation}
and $Phi(x)$ is the CDF function, any ideas?
integration normal-distribution
asked Nov 22 at 5:11
Diana
112
I'm trying to evaluate the integral
begin{equation}
int_{-infty}^{infty} s^2 f_{s_1} (s) Phi Big(dfrac{s}{sigma_n} Big) , ds.
end{equation}
where begin{equation}
f_{bar{s} _l}(s) = dfrac{1}{sqrt{2pi sigma_{1}^2}} e^{tfrac{text{-}(s-{s}_l)^2}{2sigma_1^2}},
end{equation}
and $Phi(x)$ is the CDF function, any ideas?
integration normal-distribution
integration normal-distribution
asked Nov 22 at 5:11
Diana
112
asked Nov 22 at 5:11
Diana
112
edited Nov 23 at 8:46
asked Nov 22 at 5:11
Diana
112
asked Nov 22 at 5:11
Diana
112
asked Nov 22 at 5:11
Diana
112
112
Not indefinite, but improper integral.
– StubbornAtom
Nov 22 at 5:37
add a comment |
Not indefinite, but improper integral.
– StubbornAtom
Nov 22 at 5:37
Not indefinite, but improper integral.
– StubbornAtom
Nov 22 at 5:37
Not indefinite, but improper integral.
– StubbornAtom
Nov 22 at 5:37
add a comment |
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Not indefinite, but improper integral.
– StubbornAtom
Nov 22 at 5:37