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MGF(Moment Generating Function)

Example 1:
Let $y_1 … y_2$ be independent $Expo(\beta)\ r.v\ ‘s$. What is the dirstribution of $x = ay_1+ay_2+…ay_n$?
MGF of x is $M_x(t) = E[e^ {(tx)}] = E[e^{t(ay_1+ay_2+…ay_n)}]$
$= E[e^{tay_1}e^{tay_2}…e^{tay_n}] \overset{ {\text{indep}}}{=} E[e^{tay_1}]E[e^{tay_2}]…E[e^{tay_n}]$
$= M_{y_{1}}(at)M_{y_{2}}(at)…M_{y_{n}}(at)$
$= (\frac{1}{1-\beta at})^n$
This is the MGF of $Gamma(n, \beta a)$
$\Rightarrow x \sim Gamma(n, \beta a)$ since MGF determines the distribution.

Example 2:
Let $f(x,y)$be the joint PDF for r.v.s X and Y, then the marginal PDF of x is $F_x(x) = \int_{-\infty}^{+\infty}f(x,y)dy$; the marginal PDF of y is $F_y(y) = \int_{-\infty}^{+\infty}f(x,y)dx$.

• If x and y are independent, then $f(x,y) = f_x(x)\cdot f_y(y)$
• If a joint density F(x,y) can be factored so that $f(x,y) = g(x)h(y)$, then x, y are independet and $g(x)\ and\ h(y)$are only function of x and y respectively.

Bivariate transformation

Suppose x, y are joint continuous with joint PDF $f_{x,y}(x,y)$, consider a 1-1, onto transformation, $U = g_1(x,y),\ V = g_2(x,y)$, with the inverse transformation, we can find:
$$X = h_1(U,V)$$
$$Y = h_2(U,V)$$
Then we need to get the jacobian of the transformation:
$$J = det\begin{bmatrix} \frac{\alpha x}{\alpha U} & \frac{\alpha x}{\alpha V} \\ \frac{\alpha y}{\alpha U} & \frac{\alpha y}{\alpha V} \end{bmatrix} = \frac{\alpha x}{\alpha U} \frac{\alpha y}{\alpha V} - \frac{\alpha x}{\alpha V} \frac{\alpha y}{\alpha U}$$
Therefore, $f_{U,V}(U,V) = f_{x,y}(h_1(U,V), h_2(U,V))|J|$ for all (U,V) where $f_{x,y}(x,y) > 0$.

Example 1:
Let X, Y be independent $N\sim (0,1)$. Define $U = x+y,\ V=x-y$. Find joint PDF of U and V.
Joint PDF: $f_{x,y}(x,y)\overset {\text{indep}}{=} \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} \frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}$, where $x \in \mathbb{R},\ y\in \mathbb{R}$.
The inverse transformation are:
$$U+V=2x\Rightarrow x=\frac{U+V}{2}=h_1(U,V)$$
$$U-V=2y\Rightarrow y=\frac{U-V}{2}=h_2(U,V)$$
$$\Rightarrow J = def \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix}$$
Then $f_{U,V}(U,V) = f_{x,y}{(\frac{U+V}{2},\frac{U-V}{2})}\cdot {\frac{1}{2}}$
$=\frac{1}{\sqrt {2\pi }}e^{-({\frac {U+V}{2}})^2/2}\cdot \frac{1}{\sqrt{2\pi}}e^{-({\frac{U-V}{2}})^2/2}\cdot\frac{1}{2}$
$= \frac{1}{4\pi}e^{-\frac{(U+V)^2}{8}-\frac{(U-V)^2}{8}} = \frac{1}{4\pi}exp(-\frac{1}{8}(U^2+2UV+V^2+U^2-2UV+V^2))$
$=\frac{1}{4\pi}e^{-\frac{1}{4}(U^2+V^2)} = \frac{1}{4\pi}e^{-\frac{U^2}{4}}e^{-\frac{V^2}{4}}$
$= \frac{1}{\sqrt{2\pi}\sqrt{2}}e^{-\frac{U^2}{4}}\cdot \frac{1}{\sqrt{2\pi}\sqrt{2}}e^{-\frac{V^2}{4}}$
$\Rightarrow$ then the joint PDF of U, V factors into g(U)h(V), so U,V are independent and are N(0,2).

Example 2:
Let $x \sim N(0,1)$, and $y\sim |Z|$ where $Z\sim N(0,1)$ and x, y are independent. Define $U = \frac{x}{y}, V = y, f_x(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}, x \in \mathbb{R}; f_y(y) = \frac{2}{\sqrt{2\pi}}e^{\frac{-y^2}{2}} for\ y > 0$.
So x, y have joint PDF $f_{x,y}(x,y)\overset{\text{indep}}{=}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\frac{2}{\sqrt{2\pi}}e^{\frac{-y^2}{2}}, x\in \mathbb{R}, y > 0.$
Range of transformation $U\in \mathbb{R}, V > 0$.
Get the inverse transformation: $Y = V,\ X = U\cdot V$, and jacobian is $$J = det\begin{bmatrix} V & U \\ 0 & 1 \end{bmatrix}$$
$\Rightarrow J= V\cdot 1 - U\cdot 0 = V$
So$f_{U,V}(U,V) = f_{x,y}(UV, V)|V|$
$=\frac{1}{\sqrt{2\pi}}e^{-\frac{(UV)^2}{2}}\frac{2}{\sqrt{2\pi}}e^{\frac{-V^2}{2}}$
$=\frac{V}{\pi}e^{\frac{-V^2(U^2+1)}{2}}\ for\ U\in \mathbb{R},\ V>0$
Therefore U and V are not independent because they can’t factor into $g(U)h(V)$.
The marginal PDF of U is $f_U(U) = \int_0^{\infty}f_{U,V}(U,V)dV$
$=\int_{0}^{\infty}\frac{V}{\pi}e^{\frac{-V^2(U^2+1)}{2}}dV = \int_{0}^{\infty}\frac{1}{\pi}e^{\frac{-w(U^2+1)}{2}}\frac{dw}{2}$
$=\frac{1}{2\pi}\frac{1}{\frac{-(U^2+1)}{2}}e^{\frac{-w(U^2+1)}{2}}|_0^{\infty} = \frac{1}{\pi(U^2+1)}, U\in \mathbb{R}$, where $w = V^2$, $dw = 2VdV$ and $\int e^{cw}dw = \frac{1}{c}e^{cw}$
This distribution looks like N(0,1) but has fatter tail, which go to 0 slower than N(0,1) does.
In fact E(U), Var(U) are undefined given integral doesn’t converge.
$f_U(U) = \frac{1}{\pi(U^2+1)}, U \in \mathbb{R}$ is called the cauchy distribution. Due to the symmetry around 0, we can express $\frac{Z_1}{Z_2} \sim\ Cauchy$ where $Z_1, Z_2$ are independent N(0,1).

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