Joint pdf
| |
\[
\begin{aligned}
\int \int f_{X,Y}(x,y) \, dx \, dy &= 1 \\
F_{X,Y}(x,y) &= P(X \le x, Y \le y) = \int_{-\infty}^y \int_{-\infty}^x f_{X,Y}(s,t) \,ds\, dt & \text{cumulative density function (CDF) } \\
f_{X,Y}(x,y) &= \frac{d^2}{dx\, dy} F_{X,Y}(x,y) \\
\end{aligned}
\] |
From joint to marginal
| |
\[
\begin{aligned}
f_X(x) &= \int f_{X,Y}(x,y)\,dy \\
F_X(x) &= P(X \le x) = \int_{-\infty}^x \bigg( \int_{-\infty}^\infty f_{X,Y}(s,t) \, dt \bigg) ds \\
\end{aligned}
\] |
\(Y = aX + b\)
| |
\[
\begin{aligned}
p_Y(y) &= \mathbb{P}(Y=y) = \mathbb{P}(y=aX+b) = \mathbb{P}(X=\frac{y-b}{a}) = p_X(\frac{y-b}{a}) & \text{discrete r.v.} \\
F_Y(y) &= \mathbb{P}(Y \le y) = \mathbb{P}(aX+b \le y) = \mathbb{P}(X \le \frac{y-b}{a}) = F_X(\frac{y-b}{a}) \\
\color{orange}{f_Y(y)} &= \frac{d}{dx}F_X(\frac{y-b}{a}) = \color{orange}{\frac{1}{|a|}\cdot f_X(\frac{y-b}{a})} & \text{continuous r.v.} \\
\end{aligned}
\] |
\(\color{red}{Z = X + Y}\)
\(\quad(X,Y\) independent)
| |
\[
\begin{aligned}
\mathbb{P}_Z(z) &= \sum_x \mathbb{P}(X=x, Y=z-x) = p_Z(z) \\
p_Z(z) &= \sum_x p_X(x) \cdot p_Y(z - x) \\
\end{aligned}
\] |
Discrete convolution mechanics
Given \(z\) and \(Z=X+Y\), find \(p_Z(z)\) from \(p_X(x)\) and \(p_Y(y)\).
| |
\[
\begin{aligned}
\color{blue}{f_{Z|X}(z|x)} &= f_{Y+X|X}(z|x) = f_{Y+X}(z) & \text{by independence of } X,Y \\
&= \color{blue}{f_Y(z-x)} & \color{orange}{f_{X+b}(x) = f_X(x-b)}, \text{ see Lec 11}
\end{aligned}
\] |
Joint pdf of Z and X
| |
\[
\begin{aligned}
f_{X,Z}(x,z) &= f_X(x)\cdot \color{blue}{f_{Z|X}(z|x)} = f_X(x) \cdot \color{blue}{f_Y(z-x)} \\
f_{Z}(z) &= \int_{-\infty}^\infty f_{X,Z}(x,z)\,dx = \int_{-\infty}^\infty f_X(x) \cdot \color{blue}{f_Y(z-x)} \,dx \\
\end{aligned}
\] |
\(Y = g(X)\)
\(\quad\) How to find \(f_Y(y)\) in general ?
| |
\[
\begin{aligned}
F_Y(y) &= \mathbb{P}(g(X) \le y) \\
f_Y(y) &= \frac{d}{dy}F_Y(y) \\
\end{aligned}
\] |
\(Y = g(X)\) when \(g\) is monotonic
| |
\[
\begin{aligned}
F_Y(y) &= \mathbb{P}(g(X) \le y) = \mathbb{P}(X \le h(y)) = F_X(h(y)) \\
f_Y(y) &= \frac{d}{dy}F_Y(y) = \frac{d}{dy}F_X\left(h(y)\right) = f_X\left(h(y)\right) \big\lvert \frac{d}{dy} h(y) \big\rvert \\
\end{aligned}
\] |
Source: MITx 6.041x, Lecture 9, 11, 12.
# posted by rot13(Unafba Pune) @ 4:56 PM
