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

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

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

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?

• 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. – Felix Marin Oct 15 '13 at 7:47
• 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 ...) – Jiew Meng Oct 15 '13 at 8:24
• @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 $0\mbox{k}.$ – Felix Marin Oct 15 '13 at 9:27
$$\ldots=\sqrt{1-\rho^2}\left[y\rho\sqrt{1-\rho^2}\underbrace{\int_{\mathbb{R}}ue^{-u^2/2}du}_{0\mbox{ odd integrant}}+y\rho{\int_{\mathbb{R}}e^{-u^2/2}du}\right]=y\rho\sqrt{1-\rho^2}\int_\mathbb{R}e^{-u^2/2}du=\ldots$$
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 $$\int_{\mathbb R} (2\pi)^{-1/2} u \exp(-u^2/2)\, du = 0$$