Aleatory gaussian variables

[nicola de rosa]

Let $X=N(mu_x,sigma_x^2)$ and $Y=N(mu_y,sigma_y^2)$ be two aleatory gaussian independent variables.
1)Prove that $mu_(x,y)$ is the mean and $sigma_(x,y)^2$ is the variance of $X$ and $Y$;
2)Prove that $Z=X+Y=N(mu_x+mu_y,sigma_x^2+sigma_y^2)$
3)Compute the moments of every order of a gaussian aleatory variable, standard and not.

[nicola de rosa]

does nobody try?

[nicola de rosa]

nothing?

[nicola de rosa]

Since nobody has tried to give the solution, then I give mine:
1) I show that $mu_(x,y)$ is the mean and $sigma_(x,y)^2$ is the variance of $X$ and $Y$:
The mean is: $E[X]=int_{-infty}^{+infty}xf_X(x)dx=int_{-infty}^{+infty}(x*1/sqrt(2pisigma_x^2)*e^(-(x-mu_x)^2/(2sigma_x^2)))dx$=
$int_{-infty}^{+infty}(x-mu_x)*1/sqrt(2pisigma_x^2)*e^(-(x-mu_x)^2/(2sigma_x^2))dx+mu_x*int_{-infty}^{+infty}f_X(x)dx$
Now the first integral is null because it is the integral of an odd function in a symmetrical interval respect to the origin, while the second integral gives as result $mu_x$ because $int_{-infty}^{+infty}f_X(x)dx=1$ for the normalization property of a pdf. Therefore $E[X]=mu_x$
$Var[X]=int_{-infty}^{+infty}(x-mu_x)^2f_X(x)dx=int_{-infty}^{+infty}(x-mu_x)^2*1/sqrt(2pisigma_x^2)*e^(-(x-mu_x)^2/(2sigma_x^2))dx$
We effect the substitution $t=(x-mu_x)/(sigma_x)$, obtaining
$Var[X]=sigma_x^2*int_{-infty}^{+infty}t^2*1/sqrt(2pi)*e^(-t^2/2)dt=sigma_x^2*int_{-infty}^{+infty}t*t*1/sqrt(2pi)*e^(-t^2/2)dt$=
$sigma_x^2*[-t*1/sqrt(2pi)*e^(-t^2/2)]_{-infty}^{+infty}+sigma_x^2*int_{-infty}^{+infty}1/sqrt(2pi)*e^(-t^2/2)dt=sigma_x^2$
because the first integral is null and $int_{-infty}^{+infty}1/sqrt(2pi)*e^(-t^2/2)dt=1$. So $Var[X]=sigma_x^2$
2)I prove that $Z=X+Y$, $X$ and $Y$ independent gaussian variables, is gaussian:
We must remember that the pdf of a sum of two aleatory independent variables is the convolution of pdfs and we remember that the convolution in a domain corresponds to the product in a transformed domain. For this reason it is useful to use the characteristic function (FC) which is, except for a sign to the exponent, Fourier's transform of a pdf.
Thus $FC[Z]=FC[X]*FC[Y]$ where $FC[X]=E[e^(iomega*X)]=int_{-infty}^{+infty}e^(iomega*x)f_X(x)dx$=
$int_{-infty}^{+infty}1/sqrt(2pisigma_x^2)*e^(-(x-mu_x)^2/(2sigma_x^2)+iomega*x)dx$=
$int_{-infty}^{+infty}1/sqrt(2pisigma_x^2)*e^(-((x-mu_x)/(sqrt(2)sigma_x)-(i*sigma_x*omega)/sqrt(2))^2)*e^(i*omega*mu_x-(sigma_x^2*omega^2)/2)dx$=
$e^(i*omega*mu_x-(sigma_x^2*omega^2)/2)*int_{-infty}^{+infty}1/sqrt(2pisigma_x^2)*e^(-((x-mu_x)/(sqrt(2)sigma_x)-(i*sigma_x*omega)/sqrt(2))^2)dx$=$e^(i*omega*mu_x-(sigma_x^2*omega^2)/2)$ because $int_{-infty}^{+infty}1/sqrt(2pisigma_x^2)*e^(-((x-mu_x)/(sqrt(2)sigma_x)-(i*sigma_x*omega)/sqrt(2))^2)dx=1$, being an integral of a gaussian pdf (with complex mean ).
Thus $FC[Y]=e^(i*omega*mu_y-(sigma_y^2*omega^2)/2)$, and therefore
$FC[Z]=FC[X]*FC[Y]=e^(i*omega*(mu_x+mu_y)-((sigma_x^2+sigma_y^2)*omega^2)/2)$ and presents the same form of a $FC[X]$ and $FC[Y]$ and for this reason, if we do the inverse transform, we get $Z=N(mu_x+mu_y,sigma_x^2+sigma_y^2)$
But we could obtain all what without using the properties of Fourier's transform and applying simply the definition for which $FC[Z]=E[e^(i*omega*Z)]=E[e^(i*omega*(X+Y))]=E[e^(i*omega*X)]*E[e^(i*omega*Y)]$ in which we have exploited the independence of $X$ and $Y$.
3) To find the moments of every order we exploit the following identities:
$int_{-infty}^{+infty}e^(-alpha*x^2)dx=sqrt(pi/alpha)$,
$int_{-infty}^{+infty}x^(2n)e^(-alpha*x^2)dx=(2n-1)!!*sqrt(pi/alpha)*(2alpha)^(-n)$ where the second identity can be proved exploiting the integration per parti while the first identity derives from the famous result $int_{-infty}^{+infty}e^(-x^2)dx=sqrt(pi)$ and through the substitution $t=xsqrt(alpha)$ we find the desired result . Thus if we use in the second formula $alpha=1/2$ we find:
$1/sqrt(2pi)*int_{-infty}^{+infty}x^(2n)e^(-x^2/2)dx=(2n-1)!!$ and $1/sqrt(2pi)*int_{-infty}^{+infty}x^(2n)e^(-x^2/2)dx=E[X^(2n)]$ with $X=N(0,1)$. Therefore we have showed that
for $X=N(0,1)$ $E[X^(2n)]=(2n-1)!!$ where the symbol $!!$ indicates the factorial of the odds. Obviously the moments of odd order are null because $E[X^(2n-1)]=0$ because it is the integral of an odd function (because $x^(2n-1)$ is odd and $f_X(x)$ is even, thus their product is an odd function) in a symmetrical interval respect to the origin.
For example $E[X^2]=1$ and $E[X^4]=3$.
Now if we have a gaussian non standard, that is $Y=N(mu,sigma^2)$, we can write it as $Y=sigma*X+mu$ where $X=N(0,1)$. Thus $E[Y^n]$=$E[(sigma*X+mu)^n]=E[sum_{k=0}^{n}((n),(k))sigma^k*X^k*mu^(n-k)]$=$sum_{k=0}^{n}((n),(k))sigma^kmu^(n-k)E[X^k]$
For example $E[Y]=mu$ , $E[Y^2]=sigma^2+mu^2$, $E[Y^3]=mu^3+3mu*sigma^2$ and so on.