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How do I integrate the inner integral on 2nd line?

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$$\int^\infty_{-\infty} x \exp\{ -\frac{1}{2(1-\rho^2)} (x-y\rho)^2 \} \, dx$$

I know I can use integration by substitution, let $u = \frac{x-y\rho}{\sqrt{1-\rho^2}}$ resulting in

$$\sqrt{1-\rho^2}\int^{\infty}_{-\infty} [u\sqrt{1-\rho^2} + y\rho] e^{-u^2/2} \; du$$

Thats the 3rd line in the image, but how do I proceed?

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\begin{align} &\phantom{=\,\,}\int^\infty_{-\infty}x \exp\left(% -\,\frac{\left[x - y\rho\right]^{2}}{2\left[1 - \rho^{2}\right]} \right)\,dx \tag{1} \\[3mm]&= \int^\infty_{-\infty}\left(x + y\rho\right) \exp\left(-\,\frac{x^{2}}{2\left[1 - \rho^{2}\right]}\right) \, dx \tag{2} \\[3mm]&= y\rho\,\sqrt{2\pi\,}\,\sqrt{1 - \rho^{2}\,}\ \underbrace{\quad\left[% {1 \over \sqrt{2\pi\,}\,\sqrt{1 - \rho^{2}\,}}\int^\infty_{-\infty} \exp\left(-\,\frac{x^{2}}{2(1-\rho^2)}\right) \, dx \right]\quad}_{{\LARGE =\ 1}} \tag{3} \\[3mm]&= \color{#ff0000}{\large y\rho\,\,\sqrt{\,2\pi\left(1 - \rho^{2}\right)\,\,\,}} \end{align}

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  • $\begingroup$ Looks a lot simpler but how did you get from step 1 to 2? $\endgroup$ – Jiew Meng Oct 15 '13 at 7:23
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    $\begingroup$ In $\left(2\right)$ we make a change of variable. It's like $z = x - y\rho$ but we use the same letter $x$. The piece which multiplies $x$ is zero because it is an odd function. So, you are left with $y\rho$ that goes outside the integral. In the final step, the Gaussian integral is equal to $1$. This is the usual Gaussian distribution probability. We just complete the pre factors such that the result becomes equal to 1. $\endgroup$ – Felix Marin Oct 15 '13 at 7:47
  • $\begingroup$ Err ... so you are saying $\int^\infty_{-\infty} u \exp\{-\frac{u^2}{2(1-\rho^2)}\} \, dx = 0$? Sorry ... but how do u know its an odd function? (really bad at maths ... sorry ...) $\endgroup$ – Jiew Meng Oct 15 '13 at 8:24
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    $\begingroup$ @JiewMeng $u$ is odd. ${\rm e}^{-u^{2}}$ is even. The product of both of them is odd. $\displaystyle{\int_{-\mu}^{\mu}{\rm f}\left(x\right)\,{\rm d}x = 0}$ whenever ${\rm f}$ is odd: $\displaystyle{{\rm f}\left(-x\right) = -{\rm f}\left(x\right),\ \forall\ x \in \left[-\mu,\mu\right]}$. Tomorrow, I'll will make some comment. It's too late ( 5 a.m. in the morning ). I go to sleep. Tomorrow night is $0\mbox{k}.$ $\endgroup$ – Felix Marin Oct 15 '13 at 9:27
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$$\ldots=\sqrt{1-\rho^2}\left[y\rho\sqrt{1-\rho^2}\underbrace{\int_{\mathbb{R}}ue^{-u^2/2}du}_{0\mbox{ odd integrant}}+y\rho{\int_{\mathbb{R}}e^{-u^2/2}du}\right]=y\rho\sqrt{1-\rho^2}\int_\mathbb{R}e^{-u^2/2}du=\ldots$$

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Then, recalling that $u \mapsto (2\pi)^{-1/2}\exp(-u^2/2)$ is the density of a standard normal distribution, we have that $$ \int_{\mathbb R} (2\pi)^{-1/2} \exp(-u^2/2)\, du = 1 \iff \int_{\mathbb R} \exp(-u^2/2)\, du = (2\pi)^{1/2} $$ and $\int_{\mathbb R} (2\pi)^{-1/2}u\exp(-u^2/2)\, du$ is the expectation of a standard normal, hence equal to 0 (you can also argue that this holds as the integrand is odd), so $$ \int_{\mathbb R} (2\pi)^{-1/2} u \exp(-u^2/2)\, du = 0 $$

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