Transformations and Convolutions

Theorem 5.1 (CDF Method). If $Y = g(X)$ and $g$ is monotone, then

$F_Y(y) = P(g(X) \leq y) = \begin{cases} F_X(g^{-1}(y)) & \text{if} g \text{ is} increasing \\ 1 - F_X(g^{-1}(y)) & \text{if} g \text{ is} decreasing \end{cases}$

Theorem 5.2 (Change of Variables). If $Y = g(X)$ where $g$ is differentiable and strictly monotone, then

$f_Y(y) = f_X(g^{-1}(y)) \cdot \left|\frac{d}{dy} g^{-1}(y)\right|$

Theorem 5.3. If $X$ and $Y$ are independent continuous random variables, the PDF of $Z = X + Y$ is

$f_Z(z) = (f_X * f_Y)(z) = \int_{-\infty}^{\infty} f_X(x)\, f_Y(z - x)\, dx$

Proof. $F_Z(z) = P(X + Y \leq z) = \iint_{x+y \leq z} f_{X,Y}(x, y)\, dx\, dy = \int_{-\infty}^{\infty} f_X(x)\left[\int_{-\infty}^{z-x} f_Y(y)\, dy\right] dx = \int_{-\infty}^{\infty} f_X(x)\, F_Y(z - x)\, dx$.

Differentiating: $f_Z(z) = \int_{-\infty}^{\infty} f_X(x)\, f_Y(z - x)\, dx$. $\blacksquare$

Corollary 5.4. The sum of independent normals is normal: if $X \sim N(\mu_1, \sigma_1^2)$ and $Y \sim N(\mu_2, \sigma_2^2)$ are independent, then $X + Y \sim N(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)$.

Proof. The convolution of two Gaussian PDFs is Gaussian. This follows from the MGF: $M_{X+Y}(t) = M_X(t)M_Y(t) = \exp((\mu_1 + \mu_2)t + (\sigma_1^2 + \sigma_2^2)t^2/2)$Which is the MGF of $N(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)$. $\blacksquare$

• Confusing PDF and CDF. PDF $f(x)$: probability density; CDF $F(x) = P(X \leq x) = \int_{-\infty}^x f(t)\, dt$. Fix: $F"(x) = f(x)$; $P(a < X < b) = F(b) - F(a)$.
• Wrong central limit theorem application. The CLT applies to the sample mean, not individual observations, and requires sufficiently large $n$. Fix: $\bar{X}_n \xrightarrow{d} N(\mu, \sigma^2/n)$ as $n \to \infty$.
• Confusing type I and type II errors. Type I: rejecting $H_0$ when it is true ($\alpha$). Type II: failing to reject $H_0$ when it is false ($\beta$). Fix: Type I = false positive; Type II = false negative. Decreasing one increases the other.

Problem. $X \sim N(100, 15^2)$. Find $P(X > 130)$.

Solution. $Z = \frac{130 - 100}{15} = 2.0$. $P(X > 130) = P(Z > 2) = 1 - \Phi(2) \approx 1 - 0.9772 = 0.0228$.

$\blacksquare$

Problem. Test $H_0: \mu = 50$ vs $H_1: \mu > 50$ given $\bar{x} = 53$, $s = 8$, $n = 25$, $\alpha = 0.05$.

Solution. $t = \frac{53 - 50}{8/\sqrt{25}} = \frac{3}{1.6} = 1.875$. Critical value: $t_{0.05, 24} = 1.711$. Since $1.875 > 1.711$, reject $H_0$ at the 5% level.

$\blacksquare$

• Continuous distributions: PDF integrates to 1; CDF gives cumulative probability.
• Normal distribution: $X \sim N(\mu, \sigma^2)$; standardise: $Z = (X - \mu)/\sigma$.
• Central limit theorem: sample mean is approximately normal for large $n$.
• Hypothesis testing: state $H_0$ and $H_1$, choose significance level, compute test statistic, compare with critical value.

Convolution satisfies several algebraic properties that simplify calculations:

Commutativity: $f_X * f_Y = f_Y * f_X$. The order of summation does not matter.

Associativity: $(f_X * f_Y) * f_Z = f_X * (f_Y * f_Z)$. Multiple sums can be evaluated in any order.

Scaling: If $Z = aX$ where $a > 0$, then $f_Z(z) = \frac{1}{a}f_X(z/a)$.

Location-scale family: If $X$ has PDF $f(x)$, then $Y = aX + b$ has PDF $f_Y(y) = \frac{1}{a}f\!\left(\frac{y - b}{a}\right)$.

The moment generating function (MGF) is $M_X(t) = E[e^{tX}]$.

Key property: If $M_X(t) = M_Y(t)$ for all $t$ in a neighbourhood of 0, then $X$ and $Y$ have the same distribution (MGFs uniquely determine distributions).

Linear transformations: If $Y = aX + b$, then $M_Y(t) = e^{bt}M_X(at)$.

Sums of independent variables: If $X$ and $Y$ are independent, $M_{X+Y}(t) = M_X(t)\cdot M_Y(t)$.