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.

$\blacksquare$

flowchart TD
    A[5_Transformations And Convolutions] --> B[Key Definitions]
    A --> C[Main Theorems]
    A --> D[Applications]
    B --> E[Core concepts]
    C --> F[Proof techniques]
    D --> G[Problem solving]

• 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.

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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, $M_{X+Y}(t) = M_X(t)\cdot M_Y(t)$.

When you transform a random variable, its distribution changes according to how the transformation stretches or compresses the number line. The change-of-variables formula accounts for the Jacobian — the local stretching factor. When you add two independent random variables, their distributions convolve: the density of the sum at a point is the integral of one density times a shifted version of the other, like sliding two waveforms past each other. Moment generating functions turn this convolution into multiplication, making sums of independent variables easy to handle. The central limit theorem is the grand payoff: no matter what distribution you start with, the sum of many independent copies approaches a normal distribution because convolutions smooth out irregularities.

• Probability Theory
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