PDF, CDF, expectation, transformations, and convolutions
A random variable is continuous if it can take any value in an interval. Individual values have probability zero — probability is only meaningful for intervals. The PDF plays the role of the PMF, but it's a density, not a probability.
A random variable \(X\) is continuous with PDF \(f_X(x)\) if:
\[\mathbb{P}(a \leq X \leq b) = \int_a^b f_X(x)\,dx\]
Valid PDFs satisfy: \(f_X(x) \geq 0\) and \(\int_{-\infty}^{\infty} f_X(x)\,dx = 1\).
\[F_X(x) = \mathbb{P}(X \leq x) = \int_{-\infty}^x f_X(t)\,dt\]
The PDF is the derivative of the CDF: \(f_X(x) = F_X'(x)\).
Linearity of expectation holds: \(\mathbb{E}[aX+b] = a\mathbb{E}[X]+b\). Properties of variance: \(\text{Var}(aX+b) = a^2\text{Var}(X)\).
If \(Y = g(X)\) where \(g\) is strictly monotone and differentiable:
\[f_Y(y) = f_X(g^{-1}(y))\left|\frac{d}{dy}g^{-1}(y)\right|\]The absolute value of the derivative (Jacobian) accounts for stretching/compressing of probability mass under the transformation.
Given the joint PDF \(f_{X,Y}(x,y)\) and a transformation \((U,V) = g(X,Y)\), the joint PDF of \((U,V)\) is:
\[f_{U,V}(u,v) = f_{X,Y}(x,y) \left|\det\frac{\partial(x,y)}{\partial(u,v)}\right|\]The Jacobian matrix \(\frac{\partial(x,y)}{\partial(u,v)}\) has entries \(\partial x/\partial u, \partial x/\partial v, \partial y/\partial u, \partial y/\partial v\). In 2D: \(|\det J| = |ad-bc|\) for a 2×2 matrix.
If \(X\) and \(Y\) are independent continuous random variables, the PDF of \(Z = X + Y\) is the convolution of their PDFs:
\[f_{X+Y}(t) = \int_{-\infty}^{\infty} f_X(x)\, f_Y(t-x)\, dx\]
Example: If \(X, Y \sim \mathcal{N}(0,1)\) i.i.d., then \(f_{X+Y}(t) = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-x^2/2} \cdot \frac{1}{\sqrt{2\pi}}e^{-(t-x)^2/2}\,dx\). Completing the square shows \(X+Y \sim \mathcal{N}(0,2)\).