1.4 The gaussian random variable

The random variable $X$ is said to be a Gaussian random variable

No closed form expression exists for the probability distribution function of a Gaussian random variable. For azero-mean, unit-variance, Gaussian random variable $(0, 1)$ , the probability that it exceeds the value $x$ is denoted by $Q(x)$ . $$(X> x)=1-P(X, x)=\frac{1}{\sqrt{2\pi }}\int_x \left\{x\right\}\,d$$∞ 2 2 Q x

We will have occasion to evaluate the expected value of $e^{aX+bX^2}$ where $(X, (m, ^2))$ and $a$ , $b$ are constants. By definition, $$(e^{aX+bX^2})=\frac{1}{\sqrt{2\pi ^2}}\int_{()} \,d x$$∞ ∞ a x b x 2 x m 2 2 2 The argument of the exponential requires manipulation (i.e., completing the square) before the integral can be evaluated.This expression can be written as $$-(\frac{1}{2^2}((1-2b^2)x^2-2(m+a^2)x+m^2))$$ Completing the square, this expression can be written $$-(\frac{1-2b^2}{2^2}(x-\frac{m+a^2}{1-2b^2})^2)+\frac{1-2b^2}{2^2}\left(\frac{m+a^2}{1-2b^2}\right)^2-\frac{m^2}{2^2}$$ We are now ready to evaluate the integral. Using this expression, $$(e^{aX+bX^2})=e^{\frac{1-2b^2}{2^2}\left(\frac{m+a^2}{1-2b^2}\right)^2-\frac{m^2}{2^2}}\frac{1}{\sqrt{2\pi ^2}}\int_{()} \,d x$$∞ ∞ 1 2 b 2 2 2 x m a 2 1 2 b 2 2 Let $$=\frac{x-\frac{m+a^2}{1-2b^2}}{\frac{}{\sqrt{1-2b^2}}}$$ which implies that we must require that $1-2b^2> 0$ (or $b< \frac{1}{2^2}$ ). We then obtain $$(e^{aX+bX^2})=e^{\frac{1-2b^2}{2^2}\left(\frac{m+a^2}{1-2b^2}\right)^2-\frac{m^2}{2^2}}\frac{1}{\sqrt{1-2b^2}}\frac{1}{\sqrt{2\pi }}\int_{()} \,d$$∞ ∞ 2 2 The integral equals unity, leaving the result

• $a=0$ , $(X, (m, ^2))$ $$(e^{bX^2})=\frac{e^{\frac{bm^2}{1-2b^2}}}{\sqrt{1-2b^2}}$$
• $a=0$ , $(X, (0, ^2))$ $$(e^{bX^2})=\frac{1}{\sqrt{1-2b^2}}$$
• $(X, (0, ^2))$ $$(e^{aX+bX^2})=\frac{e^{\frac{a^2^2}{2(1-2b^2)}}}{1-2b^2}$$

The real-valued random vector $X$ is said to be a Gaussian random vector if its joint distribution function has the form