Question

Show that $Z$is a standard normal random variable and if $Y$is defined by $Y=a+bZ+c{Z}^2$, then

$\rho (Y,Z)=\frac{b}{\sqrt{b^2+2c^2}}$

Short answer

We prove that$Y=a+bZ+c{Z}^2$

Step-by-step solution

Given information

Given in the question that,$Y=a+bZ+c{Z}^2$.

Explanation

Let $Z$is a standard normal random variable $N(0,1)$, and $Y=a+bZ+c{Z}^2$

Now, it can be shown$\rho (Y,Z)$

$\rho (Y,Z)=\frac{Cov(Y,Z)}{\sqrt{Var\left(Y\right)}\sqrt{Var\left(Z\right)}}$

$=bVar\left(Z\right)+cE\left(Z^3\right)-cE\left(Z^2\right)E\left(Z\right)$

$E\left(Z\right)=0,Var\left(Z\right)=1$, because $Z$are standard normal random variable.

$E\left(Z^3\right)=0$order moments are always zero,

$Cov(Y,Z)=bVar\left(Z\right)=b$

$Var\left(Y\right)=b^{2}Var\left(Z\right)+c^{2}Var\left(Z^2\right)$

$=b^2+c^{2}E\left(Z^4\right)-c^{2}E{\left(Z^2\right)}^2$

$=b^2+3c^2-1c^2$

$=b^2+2c^2$

$\rho (Y,Z)=\frac{b}{\sqrt{b^2+2c^2}\sqrt{1}}=\frac{b}{\sqrt{b^2+2c^2}}$

Hence, we obtain:

$\rho (Y,Z)=\frac{b}{\sqrt{b^2+2c^2}}$

Final answer

We prove that$Y=a+bZ+c{Z}^2$

and

$\rho (Y,Z)=\frac{b}{\sqrt{b^2+2c^2}}$