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Why Functions of Random Variables?

In practice, the quantity we observe is rarely the "raw" random variable — it is some function of it. A detector squares the received signal; an amplifier clips it; a phase estimator applies the arctangent. To analyze these systems, we need to derive the distribution of $Y = g(X)$ from the known distribution of $X$ . The tools in this section — the CDF method and the change-of-variables formula — are the workhorses for this task.

Theorem: The CDF Method

Let $X$ be a continuous RV with CDF $F$ and let $Y = g(X)$ for a Borel-measurable $g$ . Then the CDF of $Y$ is

$F_Y(y) = \mathbb{P}(g(X) \leq y) = \mathbb{P}(X \in B_g(y)),$

where $B_g(y) = \{x : g(x) \leq y\}$ . For continuous $X$ :

$F_Y(y) = \int_{B_g(y)} f(x)\,dx.$

The PDF of $Y$ (when it exists) is obtained by differentiating: $f_Y(y) = F_Y'(y)$ .

The CDF method is completely general — it works for any measurable $g$ , monotone or not. The procedure is: (1) express the event $\{Y \leq y\}$ in terms of $X$ , (2) integrate the density of $X$ over the resulting region, (3) differentiate in $y$ if you want the PDF.

Proof

Direct from the definition

$F_Y(y) = \mathbb{P}(Y \leq y) = \mathbb{P}(g(X) \leq y) = \mathbb{P}(X \in \{x : g(x) \leq y\}) = \int_{\{x : g(x) \leq y\}} f(x)\,dx$ . $\blacksquare$

Theorem: Change of Variables for Monotonic Transformations

Let $X$ be a continuous RV with PDF $f$ and let $g$ be a strictly monotonic, differentiable function with inverse $g^{-1}$ . Then $Y = g(X)$ has PDF

$f_Y(y) = f(g^{-1}(y))\,\left|\frac{d}{dy}g^{-1}(y)\right|.$

Proof

Case 1: $g$ strictly increasing

$F_Y(y) = \mathbb{P}(g(X) \leq y) = \mathbb{P}(X \leq g^{-1}(y)) = F(g^{-1}(y))$ . Differentiating: $f_Y(y) = f(g^{-1}(y))\cdot(g^{-1})'(y)$ .

Case 2: $g$ strictly decreasing

$F_Y(y) = \mathbb{P}(g(X) \leq y) = \mathbb{P}(X \geq g^{-1}(y)) = 1 - F(g^{-1}(y))$ . Differentiating: $f_Y(y) = -f(g^{-1}(y))\cdot(g^{-1})'(y) = f(g^{-1}(y))\cdot|(g^{-1})'(y)|$ , since $(g^{-1})'(y) < 0$ . Both cases yield the same formula with the absolute value. $\blacksquare$

Theorem: Non-Monotonic Transformations

Let $g : \mathbb{R} \to \mathbb{R}$ be piecewise monotonic. Partition $\mathbb{R}$ into intervals $A_1, \ldots, A_m$ on each of which $g$ is strictly monotonic with inverse $g_k^{-1}$ . Then $Y = g(X)$ has PDF

$f_Y(y) = \sum_{k=1}^{m} f(g_k^{-1}(y))\,\left|\frac{d}{dy}g_k^{-1}(y)\right|\,\mathbf{1}_{\{y \in g(A_k)\}}.$

Proof

Partition and sum

$\{g(X) \leq y\} = \bigcup_{k=1}^m \{X \in A_k,\,g(X) \leq y\}$ . These events are disjoint. On each $A_k$ , $g$ is monotonic, so we apply the previous theorem to each piece and sum. $\blacksquare$

Example: Squared Gaussian $\to$ Chi-Squared

Let $Z \sim \mathcal{N}(0, 1)$ and $Y = Z^2$ . Derive the PDF of $Y$ .

Solution

CDF method

For $y \geq 0$ : $F_Y(y) = \mathbb{P}(Z^2 \leq y) = \mathbb{P}(-\sqrt{y} \leq Z \leq \sqrt{y}) = \Phi(\sqrt{y}) - \Phi(-\sqrt{y}) = 2\Phi(\sqrt{y}) - 1$ .

Differentiate

$f_Y(y) = 2\phi(\sqrt{y})\cdot\frac{1}{2\sqrt{y}} = \frac{1}{\sqrt{y}}\cdot\frac{1}{\sqrt{2\pi}}e^{-y/2} = \frac{1}{\sqrt{2\pi y}}\,e^{-y/2}$ for $y > 0$ .

Identify the distribution

This is the $\text{Gamma}(1/2, 1/2)$ distribution, also known as the chi-squared distribution with 1 degree of freedom, $\chi^2_1$ .

Example: Complex Gaussian Envelope $\to$ Rayleigh

Let $Z = X + jY$ where $X, Y \sim \mathcal{N}(0, \sigma^2)$ are independent. Derive the distribution of the envelope $R = |Z| = \sqrt{X^2 + Y^2}$ .

Solution

CDF of $R$

$F_R(r) = \mathbb{P}(X^2 + Y^2 \leq r^2) = \int\!\!\int_{x^2+y^2 \leq r^2} \frac{1}{2\pi\sigma^2}e^{-(x^2+y^2)/(2\sigma^2)}\,dx\,dy$ .

Switch to polar coordinates

Let $x = \rho\cos\theta$ , $y = \rho\sin\theta$ : $F_R(r) = \int_0^{2\pi}\int_0^r \frac{\rho}{2\pi\sigma^2}e^{-\rho^2/(2\sigma^2)}\,d\rho\,d\theta = \int_0^r \frac{\rho}{\sigma^2}e^{-\rho^2/(2\sigma^2)}\,d\rho$ .

Differentiate to get PDF

$f_R(r) = \frac{r}{\sigma^2}\exp\!\left(-\frac{r^2}{2\sigma^2}\right)$ for $r \geq 0$ . This is the Rayleigh distribution with parameter $\sigma$ .

Example: Squared Envelope $\to$ Exponential

With $R$ Rayleigh as above, show that $W = R^2 = |Z|^2$ is exponentially distributed.

Solution

CDF method

$F_W(w) = \mathbb{P}(R^2 \leq w) = \mathbb{P}(R \leq \sqrt{w}) = F_R(\sqrt{w})$ $= 1 - e^{-w/(2\sigma^2)}$ for $w \geq 0$ .

Identify

This is the CDF of $\text{Exp}(1/(2\sigma^2))$ . The squared envelope of a complex Gaussian signal is exponentially distributed. This fact underpins the analysis of Rayleigh fading channels, where the instantaneous received power (proportional to $|h|^2$ ) is exponential.

RV Transformation Visualizer

Visualize how the CDF method works: choose a transformation $g(x)$ and see how the PDF of $X$ maps to the PDF of $Y = g(X)$ . The shaded regions show corresponding probability masses.

Parameters

Transformation

g(x)

\mu

(input Gaussian mean)0

\sigma

(input Gaussian std)1

The CDF Method for Deriving Distributions

Input: PDF

f

of

X

, transformation

g

Output: PDF

f_Y

of

Y = g(X)

1. Write

F_Y(y) = \mathbb{P}(g(X) \leq y)

2. Express

\{g(X) \leq y\}

as

\{X \in B_g(y)\}

where

B_g(y) = \{x : g(x) \leq y\}

3. Compute

F_Y(y) = \int_{B_g(y)} f(x)\,dx

4. Differentiate:

f_Y(y) = \frac{d}{dy} F_Y(y)

5. Verify:

\int f_Y(y)\,dy = 1

Step 2 is the critical step. For monotone $g$ , $B_g(y)$ is a half-line. For non-monotone $g$ (e.g., $g(x) = x^2$ ), $B_g(y)$ may be a union of intervals.

Common Mistake: Forgetting the Jacobian

Mistake:

Writing $f_Y(y) = f(g^{-1}(y))$ without the $|(g^{-1})'(y)|$ factor.

Correction:

The correct formula is $f_Y(y) = f(g^{-1}(y))\,|(g^{-1})'(y)|$ . The Jacobian factor accounts for the "stretching" or "compression" of probability mass under the transformation. Without it, $f_Y$ will not integrate to 1.

Quick Check

If $X \sim \mathcal{N}(0, 1)$ and $Y = 3X + 5$ , what is the distribution of $Y$ ?

$\mathcal{N}(5, 3)$

$\mathcal{N}(5, 9)$

$\mathcal{N}(3, 5)$

$\mathcal{N}(15, 9)$

Correction:

\mathcal{N}(5, 9)

$Y = aX + b$ with $a=3, b=5$ . So $\mathbb{E}[Y] = 5$ and $\text{Var}(Y) = 9\cdot 1 = 9$ . The Gaussian is closed under affine transformations.

Jacobian

In the change-of-variables formula, the Jacobian $|(g^{-1})'(y)|$ is the absolute value of the derivative of the inverse transformation. It corrects for the stretching or compression of the density under the mapping $Y = g(X)$ .

Chi-Squared Distribution

The distribution of $Y = Z^2$ where $Z \sim \mathcal{N}(0,1)$ . More generally, $\chi^2_d = Z_1^2 + \cdots + Z_d^2$ with $d$ i.i.d. standard normals. It is a special case of the Gamma distribution: $\chi^2_d = \text{Gamma}(d/2, 1/2)$ .

Key Takeaway

The CDF method ( $F_Y(y) = \mathbb{P}(g(X) \leq y)$ , then differentiate) is the universal technique for deriving the distribution of a transformed random variable. For monotonic $g$ , the change-of-variables shortcut with the Jacobian is faster. For non-monotonic $g$ , partition the domain and sum contributions from each monotonic piece.

Functions of a Random Variable

Why Functions of Random Variables?

Theorem: The CDF Method

Direct from the definition

Theorem: Change of Variables for Monotonic Transformations

Case 1: $g$ strictly increasing

Case 2: $g$ strictly decreasing

Theorem: Non-Monotonic Transformations

Partition and sum

Example: Squared Gaussian →\to→ Chi-Squared

CDF method

Differentiate

Identify the distribution

Example: Complex Gaussian Envelope →\to→ Rayleigh

CDF of $R$

Switch to polar coordinates

Differentiate to get PDF

Example: Squared Envelope →\to→ Exponential

CDF method

Identify

RV Transformation Visualizer

Parameters

The CDF Method for Deriving Distributions

Common Mistake: Forgetting the Jacobian

Quick Check

Jacobian

Chi-Squared Distribution

Key Takeaway

Example: Squared Gaussian $\to$ Chi-Squared

Example: Complex Gaussian Envelope $\to$ Rayleigh

Example: Squared Envelope $\to$ Exponential