Answer to Question #320984 in Statistics and Probability for ken

Question

 Let X and Y be two independent random variables having joint probability density function f(x, y) = 1/ 2πσ2 e − (x−µ) 2 σ2 e − (y−µ) 2 σ2 − ∞ < x, y < ∞

Find the moment generating function of Z = X+Y 2 and hence the mean and variance of Z

Answer 1

$f\left( x,y \right) =\frac{1}{2\pi \sigma ^2}\exp \left( -\frac{\left( x-\mu \right) ^2}{2\sigma ^2} \right) \exp \left( -\frac{\left( y-\mu \right) ^2}{2\sigma ^2} \right) \\This\,\,is\,\,a\,\,density\,\,of\,\,i.i.d. random\,\,variables\,\,X,Y\sim N\left( \mu ,\sigma ^2 \right) \\M_Z\left( t \right) =Ee^{tZ}=E^{t\left( X+Y^2 \right)}=\\=Ee^{tX}Ee^{tY^2}=\exp \left( \mu t+\frac{1}{2}\sigma ^2t^2 \right) \int_{-\infty}^{+\infty}{\frac{1}{\sqrt{2\pi \sigma ^2}}\exp \left( ty^2-\frac{\left( y-\mu \right) ^2}{2\sigma ^2} \right)}dy=\\=\exp \left( \mu t+\frac{1}{2}\sigma ^2t^2 \right) \int_{-\infty}^{+\infty}{\frac{1}{\sqrt{2\pi \sigma ^2}}\exp \left( -\frac{\left( y-\frac{\mu}{1-2t\sigma ^2} \right) ^2}{\frac{2\sigma ^2}{1-2t\sigma ^2}} \right) \exp \left( \mu ^2t \right) dt}=\\=\left[ \begin{array}{c} f\left( t \right) =\frac{1}{\sqrt{2\pi \frac{\sigma ^2}{1-2t\sigma ^2}}}\exp \left( -\frac{\left( y-\frac{\mu}{1-2t\sigma ^2} \right)}{\frac{2\sigma ^2}{1-2t\sigma ^2}} \right) \sim N\left( \frac{\mu}{1-2t\sigma ^2},\frac{\sigma ^2}{1-2t\sigma ^2} \right) \Rightarrow\\ \Rightarrow \int_{-\infty}^{+\infty}{f\left( t \right) dt}=1\\\end{array} \right] =\\=\exp \left( \mu t+\frac{1}{2}\sigma ^2t^2 \right) \frac{\exp \left( \mu ^2t \right)}{\sqrt{1-2t\sigma ^2}}=\frac{\exp \left( \left( \mu +\mu ^2 \right) t+\frac{1}{2}\sigma ^2t^2 \right)}{\sqrt{1-2t\sigma ^2}}\\EZ=\frac{d}{dt}M_Z\left( t \right) |_{t=0}=\\=\exp \left( \left( \mu +\mu ^2 \right) t+\frac{1}{2}\sigma ^2t^2 \right) \frac{\left( -2\left( \mu +\mu ^2 \right) \sigma ^2t+\mu +\mu ^2-2\sigma ^4t^2+\sigma ^2t+\sigma ^2 \right)}{\left( 1-2t\sigma ^2 \right) ^{3/2}}|_{t=0}=\\=\mu +\mu ^2+\sigma ^2\\EZ^2=\frac{d^2}{dt^2}M_Z\left( t \right) |_{t=0}=\\=\exp \left( \left( \mu +\mu ^2 \right) t+\frac{1}{2}\sigma ^2t^2 \right) \left( \frac{3\sigma ^4}{\left( 1-2t\sigma ^2 \right) ^{5/2}}+\frac{2\sigma ^2\left( \mu +\mu ^2+\sigma ^2t \right)}{\left( 1-2t\sigma ^2 \right) ^{3/2}}+\frac{\left( \mu +\mu ^2+\sigma ^2t \right) ^2+\sigma ^2}{\left( 1-2t\sigma ^2 \right) ^{1/2}} \right) |_{t=0}=\\=3\sigma ^4+2\sigma ^2\left( \mu +\mu ^2 \right) +\left( \mu +\mu ^2 \right) ^2+\sigma ^2\\Var\left( Z \right) =3\sigma ^4+2\sigma ^2\left( \mu +\mu ^2 \right) +\left( \mu +\mu ^2 \right) ^2+\sigma ^2-\left( \mu +\mu ^2+\sigma ^2 \right) ^2=\\=\sigma ^2+2\sigma ^4$