Question

Let X be a zero mean unit variance Gaussian random variable. E[|X|] is equal to .

• A. 0.7978

Answer 1

The correct option is A 0.7978

1. a Gaussian random variable having
   
   mean = 0
   variance = 1
   
   So, p.d.f. for Gaussian random variable is rpresented mathematically as
   
   $f_x\left(x\right)=\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}$
   
   $\therefore E\left[x\right]={\int }_{-\infty }^{\infty }x{f}_x\left(x\right)dx$
   
   $\Rightarrow E\left[|x|\right]={\int }_{-\infty }^{\infty }|x|\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}dx$
   
   $\text{Now, Since}{f}_x\left(x\right)=f_x(-x)\to \text{even function}$
   
   $\text{Hence,}E\left[|x|\right]=2{\int }_0^{\infty }x\frac{1}{\sqrt{2\pi }}{e}^{\frac{-x^2}{2}}dx$
   
   $\Rightarrow E\left[|x|\right]=\frac{2}{\sqrt{2\pi }}{\int }_0^{\infty }{e}^{\frac{-x^2}{2}}dx$
   
   Let $\frac{x^2}{2}=y$
   
   at x = 0; y = 0
   
   $\Rightarrow$ xdx = dy
   
   at $x=\infty ;y=\infty$
   
   $\therefore E\left[|x|\right]=\frac{2}{\sqrt{2\pi }}{\int }_0^{\infty }{e}^{-y}dy$
   
   $\Rightarrow E\left[|x|\right]=\frac{2}{\sqrt{2\pi }}{\left[\frac{e^{-\gamma }}{-1}\right]}_0^{\infty }$
   
   $\Rightarrow E\left[|x|\right]=0.7978\times \left[\frac{e^{-\infty }-e^{-0}}{-1}\right]$
   
   $\Rightarrow E\left[|x|\right]=\mathrm{0.7978.}$