Category: Exercises

Formalising the Dirac Delta

The Dirac delta “function” $\delta$ has been a bane in my teaching since I needed to start teaching it in engineering mathematics, since it is, really, not actually a function. Many of its results seem miraculously correct, and yet we don’t have any reasonably accessible introduction to its formal definition.

We will adopt a measure-theoretic interpretation of $\delta$ and regard it as a measure. We will then define the relevant Laplace transform notions on measures that extend the classical setting. Finally, we assert that

$\displaystyle \int_{(-\infty, t]} \mathrm d\delta = U(t) \equiv \mathbb I_{[0,\infty)}(t),$

so that, in some limited sense, $U' = \delta$ .

For any measure $\mu$ on $(\mathbb R, \frak{B}(\mathbb R))$ , if the measurable function $f$ is $\mu$ -integrable, make the definition

$\displaystyle \langle \mu, f \rangle := \int_{\mathbb R} f\, \mathrm d\mu.$

whenever the right-hand side is well-defined. Using the Radon-Nikodým theorem, for suitably-defined functions $f, g$ ,

$\langle \lambda_g, f\rangle = \langle f \cdot g, \lambda \rangle = \langle \lambda_f, g \rangle,$

where the measure $\lambda_f$ is defined by $\lambda_f(S) := \int_S f\, \mathrm d\lambda$ . We remark that by the same theorem, the map $f \to \lambda_f$ is injective $\lambda$ -a.e. in the following sense: if $\lambda_f = 0$ , then $f = 0$ $\lambda$ -a.e.. Furthermore, it is bona-fide injective when restricted to the space of continuous functions (i.e. $f$ is continuous).

Define the map $K : \mathbb R^2 \to \mathbb R$ by $K(s,t) := e^{-st} \cdot U(t)$ .

Definition 1. For any measure $\mu$ on $(\mathbb R, \frak{B}(\mathbb R))$ , if $f$ is $\mu$ -integrable and $K(s, \cdot) \cdot f$ is $\mu$ -integrable, we define the $\mu$ –Laplace transform by

$\displaystyle \mathcal L_{\mu}\{f \} (s) := \langle \mu ,f \cdot K(s,\cdot)\rangle \equiv \int_{\mathbb R} K(s, \cdot) \cdot f\, \mathrm d\mu,$

whenever the right-hand side is well-defined. In particular, define $\mathcal L\{\mu\} := \mathcal L_{\mu}\{1\}$ .

Problem 1. If $K(s,\cdot)\cdot f$ is $\lambda$ -integrable, prove that $\mathcal L_{\lambda}\{f \} = \mathcal L\{f\}$ , where the right-hand side refers to the classical Laplace transform on sub-exponential functions.

(Click for Solution)

Solution. By definition, for any $s$ ,

$\begin{aligned} \mathcal L_{\lambda}\{f \}(s) &= \langle \lambda, f \cdot K(s,\cdot), \rangle \\ &= \int_{\mathbb R} K(s, \cdot) \cdot f\, \mathrm d\lambda \\ &= \int_{-\infty}^{\infty} K(s, t) \cdot f(t)\, \mathrm d \lambda( t ) \\ &= \int_{-\infty}^{\infty} e^{-st} \cdot f(t)\, \mathrm dt = \mathcal L\{f\}(s). \end{aligned}$

Hence, $\mathcal L\{\lambda_f\} \equiv \mathcal L_{\lambda_f}\{1\} \equiv \mathcal L_{\lambda}\{f\} = \mathcal L\{f\}$ .

Problem 2. Prove that the Laplace transform on measures is linear in the measures: for suitably defined measures $\mu,\nu$ , a suitably-defined function $f$ , and $\alpha \in \mathbb R$ ,

$\mathcal L_{\mu + \nu} \{f\} = \mathcal L_{\mu}\{f\} + \mathcal L_{\nu}\{f\},\quad \mathcal L_{\mu}\{\alpha f \} = \alpha \cdot \mathcal L_{\mu}\{f\}.$

Furthermore, if $\mu$ is signed, then $\mathcal L_{\alpha \mu}\{f \} = \alpha \cdot \mathcal L_{\mu}\{f\}$ .

(Click for Solution)

Solution. For additivity, we first prove the additivity of measures: for $S \in \frak B(\mathbb R)$ and suitably defined functions $f$ , we first claim that

$\begin{aligned} \langle \mu + \nu, f\rangle &= \int_{\mathbb R} f\, \mathrm d(\mu + \nu) \\ &= \int_{\mathbb R} f\, \mathrm d\mu + \int_{\mathbb R} f\, \mathrm d\nu \\ &= \langle \mu,f \rangle + \langle \nu, f\rangle, \end{aligned}$

so that for suitably chosen $s$ ,

$\begin{aligned}\mathcal L_{\mu + \nu}\{f\}(s) &= \langle \mu + \nu, K(s, \cdot) \cdot f\rangle \\ &= \langle \mu, K(s, \cdot) \cdot f \rangle + \langle \nu, K(s, \cdot) \cdot f\rangle \\ &= \mathcal L_{\mu }\{f\}(s) + \mathcal L_{ \nu}\{f\}(s) \\ &= (\mathcal L_{\mu }\{f\} + \mathcal L_{\nu }\{f\})(s).\end{aligned}$

To that end, we prove the result for the non-negative simple function $f = \sum_{i=1}^n a_i \cdot \mathbb I_{S_i}$ and leave the extensions via the monotone convergence theorem and decomposition $f = f^+ - f^-$ as an exercise:

$\begin{aligned} \int_{\mathbb R} f\, \mathrm d(\mu + \nu) &= \sum_{i=1}^n a_i (\mu + \nu)(S_i) \\ &= \sum_{i=1}^n a_i \cdot (\mu(S_i) + \nu(S_i)) \\ &= \sum_{i=1}^n a_i \cdot \mu(S_i) + \sum_{i=1}^n a_i \cdot \nu(S_i) \\ &= \int_{\mathbb R} f\, \mathrm d\mu + \int_{\mathbb R} f\, \mathrm d\nu, \end{aligned}$

as required. Scalar multiplication follows similarly.

Problem 3. Prove that for any $a \in \mathbb R$ , the map $\delta_a$ defined by $\delta_a(S) := \mathbb I_{S}(a)$ is a probability measure. We call $\delta := \delta_0$ the Dirac measure and observe that that $\delta_a = \delta ( \cdot - a)$ .

(Click for Solution)

Solution. We verify the three properties of a probability measure. Firstly,

$\delta(\emptyset) = \mathbb I_{\emptyset}(0) = 0.$

Secondly,

$\begin{aligned} \delta \left( \bigsqcup_{i=1}^\infty S_i \right) &= \mathbb I_{\bigsqcup_{i=1}^\infty S_i }(0) = \sum_{i=1}^\infty \mathbb I_{ S_i }(0) = \sum_{i=1}^\infty \delta(S_i).\end{aligned}$

Finally, $\delta(\mathbb R) = \mathbb I_{\mathbb R}(0) = 1$ .

Problem 4. Prove the sifting property:

$\displaystyle \langle \delta_a, f \rangle := \int_{\mathbb R} f\, \mathrm d\delta_a = f(a).$

Deduce that $\mathcal L_{\delta_a}\{f\}= K(\cdot, a) \cdot f$ . In particular, $\mathcal L_{\delta_a}\{1\}(s) = e^{-as}$ .

(Click for Solution)

Solution. We can use the same non-negative simple $\Rightarrow$ non-negative $\Rightarrow$ integrable technique for the sifting property, which we will carry out for completeness.

Let $f = \sum_{i=1}^n a_i \cdot \mathbb I_{S_i}$ be a non-negative simple function. We observe that

$\delta_a(S_i) = 1 \quad \iff \quad a \in S_i \quad \iff \quad f(a) = a_i.$

Then

$\begin{aligned} \int_{\mathbb R} f\, \mathrm d\delta_a &= \sum_{i=1}^n a_i \cdot \delta_a(S_i) = a_i \cdot 1 = f(a).\end{aligned}$

Now suppose $f$ is non-negative. Find a sequence $\varphi_n$ of simple functions such that $\varphi_n \to f$ monotonically. By the monotone convergence theorem,

$\begin{aligned} \int_{\mathbb R} f\, \mathrm d\delta_a &=\lim_{n \to \infty} \int_{\mathbb R} \varphi_n\, \mathrm d\delta_a = \lim_{n \to \infty} \varphi_n(a) = f(a).\end{aligned}$

Finally, suppose $f$ is integrable. Writing $f = f^+ - f^-$ ,

$\begin{aligned}\int_{\mathbb R} f\, \mathrm d\delta_a &= \int_{\mathbb R} f^+\, \mathrm d\delta_a - \int_{\mathbb R} f^-\, \mathrm d\delta_a \\ &= f^+(a) - f^-(a) \\ &= (f^+ - f^-)(a) = f(a).\end{aligned}$

In particular, for any fixed $s$ ,

$\displaystyle \mathcal L_{\delta_a}\{f\}(s) = \langle \delta_a, K(s,\cdot) \cdot f\rangle = \int_{\mathbb R} K(s, \cdot) \cdot f\ \mathrm d\delta_a = K(s, a) \cdot f(a) .$

In particular, $\mathcal L_{\delta_a}\{1\} = K(\cdot, a)$ .

Remark 1. We have shown that $\mathcal L_{\mu}\{f\}$ is a meaningful generalisation of both the usual Laplace transform $\mathcal L\{f\}$ and the Laplace transform of the measure $\mathcal L\{ {\delta_a} \}:=\mathcal L_{\delta_a}\{1\}$ :

$\begin{aligned} \mathcal L \{ \alpha \cdot \lambda_f + \beta \cdot \delta_a \} &= \mathcal L_{ \alpha \cdot \lambda_f + \beta \cdot \delta_a}\{1\} \\ &= \mathcal L_{ \alpha \cdot \lambda_f }\{1\} + \mathcal L_{ \beta \cdot \delta_a}\{1\}\\ &= \alpha \cdot \mathcal L_{\lambda_f}\{1\} + \beta \cdot \mathcal L_{\delta_a}\{1\} \\ &= \alpha \cdot \mathcal L \{\lambda_f\} + \beta \cdot \mathcal L \{\delta_a \} \\ &= \alpha \cdot \mathcal L \{f\} + \beta \cdot \mathcal L \{\delta_a \}. \end{aligned}$

In this manner and usage, we embed $f \to \lambda_f$ as a (signed-)measure, and define $f + \beta \cdot \delta_a := \lambda_f + \beta \cdot \delta_a$ via said embedding, finally permitting engineers to write the following guiltlessly:

$\begin{aligned} \mathcal L\{\alpha \cdot f + \beta \cdot \delta_a\} &\equiv \mathcal L \{ \alpha \cdot \lambda_f + \beta \cdot \delta_a \} \\ &= \alpha \cdot \mathcal L \{f\} + \beta \cdot \mathcal L \{\delta_a \}. \end{aligned}$

Problem 5. Prove that $\displaystyle \int_{(-\infty, t]} \mathrm d\delta_a = U_a(t)$ . Deduce that for $a > 0$ and valid $s$ ,

$\displaystyle \int_{\mathbb R} \mathrm d\delta_a = 1 \quad \text{and} \quad \mathcal L\{\delta_a\}(s) = s \cdot \mathcal L\{U_a\}(s),$

where $U_a = U( \cdot - a)$ .

(Click for Solution)

Solution. For the first claim,

$\begin{aligned} \int_{(-\infty, t]} \mathrm d\delta_a &= \int_{\mathbb R} \mathbb I_{(-\infty, t]}\, \mathrm d\delta_a = \delta_a((-\infty, t]) = \begin{cases} 1, & t \geq a, \\ 0, & t < a. \end{cases}\end{aligned}$

Taking $T \to \infty$ and $T > a$ in particular, by the monotonicity of measures,

$\begin{aligned} \int_{\mathbb R} \mathrm d\delta_a &= \lim_{T \to \infty} \int_{(-\infty, T]} \mathrm d\delta_a = \lim_{T \to \infty} U_a(T) = \lim_{T \to \infty} 1 = 1.\end{aligned}$

For the final result,

$\displaystyle \mathcal L\{U_a\}(s) = \frac{e^{-as}}{s} = \frac{\mathcal L\{\delta_a\}(s)}{s}.$

Remark 2. In classical formulas, if $f$ is differentiable such that $f' \cdot \mathbb I_{(-\infty, t]}$ is $\lambda$ -integrable for any $t \in \mathbb R$ , $\displaystyle \lim_{N \to \infty}f(-N) = 0$ , and has a Laplace transform, we have

$\displaystyle \int_{(-\infty, t]} f'\, \mathrm d\lambda = f(t),\quad \mathcal L\{f'\}(s) = s \cdot \mathcal L\{f\}(s) - f(0).$

Since Problems 3 and 4 agree with classical setups, engineers sloppily write $U' = \delta$ and use these results on their happy merry way, leaving mathematicians infuriated in having to clean up after their…stuff.

This measure-theoretic formalisation only helps us with the integration aspect of the Dirac delta, since measure theory is, in some sense a “completion” of integral calculus. Yet, a full understanding requires us to develop a “completion” of differential calculus—this generalisation builds on measure theory and discusses distribution theory.

We hinted at these objects via the notation $\langle \mu, f \rangle$ . The map

$\langle \mu, \cdot \rangle : \{\text{good functions}\} \to \mathbb R$

is called a distribution. In a sense, then we could define $U := \langle U, \cdot \rangle$ as a distribution (though this requires much more effort), and by defining differentiation of distributions, we get the true equality $U' = \delta$ . But all that work requires a lot more pain and suffering.

As completions, they become indispensable tools in solving many more differential equations than a usual post-secondary or even undergraduate course would present.

—Joel Kindiak, 16 Jul 25, 1150H

November 11, 2025
The Union Bound

Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space.

Problem 1. For $K_1, K_2, \dots\in \mathcal F$ , prove Boole’s inequality:

$\displaystyle \mathbb P \left( \bigcup_{i=1}^\infty K_i \right) \leq \sum_{i=1}^\infty \mathbb P(K_i).$

This result is also known as the union bound.

(Click for Solution)

Solution. Define $L_1 := K_1$ and inductively, $L_{i+1} = K_{i+1} \backslash \bigcup_{j=1}^i K_j$ . We can check that $\{L_i\}$ is pairwise mutually exclusive and $L_i \subseteq K_i$ for each $i$ . Furthermore, it is not hard to prove that

$\displaystyle \bigcup_{i=1}^\infty K_i = \bigsqcup_{i=1}^\infty L_i.$

To see this, we first note that $(\supseteq)$ is obvious. For the direction $(\subseteq)$ , fix $x \in K_i$ for some $i$ . Take $i$ to be the smallest by the well-ordering principle, so that $x \notin \bigcup_{j=1}^{i-1} K_j$ . Therefore, $x \in L_i$ , as required. Therefore,

$\displaystyle \mathbb P\left( \bigcup_{i=1}^\infty K_i\right) = \mathbb P \left( \bigsqcup_{i=1}^\infty L_i \right) = \sum_{i=1}^\infty \mathbb P(L_i) \leq \sum_{i=1}^\infty \mathbb P(K_i) .$

This is known as the disjoint union trick.

Problem 2. For $K_1, K_2, \dots, K_n \in \mathcal F$ , prove Bonferroni’s inequality.

$\displaystyle \mathbb P \left( \bigcap_{i=1}^n K_i \right) \geq \sum_{i=1}^n \mathbb P(K_i) - (n-1).$

(Click for Solution)

Solution. Firstly, setting $K_{n+i} = \emptyset$ for $i \geq 0$ ,

$\displaystyle \mathbb P \left( \bigcup_{i=1}^n K_i \right) = \mathbb P \left( \bigcup_{i=1}^\infty K_i \right) \leq \sum_{i=1}^\infty \mathbb P(K_i) = \sum_{i=1}^n\mathbb P(K_i).$

Defining $L_i := \Omega \backslash K_i$ for $i \geq 1$ , use Problem 1 to deduce

$\begin{aligned} \mathbb P \left( \bigcap_{i=1}^n K_i \right) &= \mathbb P \left( \bigcap_{i=1}^n ( \Omega \backslash L_i ) \right) = \mathbb P \left( \Omega \backslash \bigcup_{i=1}^n L_i \right) \\ &= 1 - \mathbb P \left( \bigcup_{i=1}^n L_i \right) \geq 1 - \sum_{i=1}^n \mathbb P(L_i) \geq 1 - \sum_{i=1}^n \mathbb P(\Omega \backslash K_i) \\ &\geq 1 - \sum_{i=1}^n \mathbb (1 - \mathbb P(K_i)) = \sum_{i=1}^n \mathbb P(K_i) - (n-1). \end{aligned}$

Corollary 1. We have the inequality

$\displaystyle \mathbb P \left( \bigcup_{i=1}^n K_i \right) \leq \sum_{i=1}^n \mathbb P(K_i) \leq \mathbb P \left( \bigcap_{i=1}^n K_i \right) + (n-1).$

—Joel Kindiak, 30 Aug 25, 2048H

October 8, 2025
The Friendship Formula
Friendship is a massively important topic of contemplation in my life, so much to the point I have proposed an equation to quantify the closeness between two friends. Suppose Persons X and Y are friends, and Person X wants to evaluate his closeness with Person Y.

Definition 1. Define the following random variables bounded in $[0, 1]$ :
- $Q$ denotes the space for vulnerability that Person X gives to Person Y (i.e. the extent to which Person Y does not need to feel defensive when interacting with Person X).
- $R$ denotes the space for vulnerability that Person X receives from Person Y.
- $S$ denotes the access that Person X gives to Person Y (i.e. the amount of personal information that Person X discloses to Person Y).
- $T$ denotes the access that Person X receives from Person Y.
Assume the random variables $Q,R,S,T$ are jointly independent continuous random variables.

Note that Person X determines these quantities, and may differ from Person Y’s evaluation (explaining why X can feel close to Y but the feelings are not mutual).

Axiom 1. The mutual closeness $C$ that X feels he shares with Y is defined by the equation

$C := \min\{ Q, R \} \cdot \min\{ S, T \}.$

Problem 1. Prove that $\mathbb E[C]$ is given by the quantity

$\displaystyle \begin{aligned} \left(\mathbb E[Q] + \mathbb E[R] - 1 + \int_{[0, 1]} F_Q \cdot F_R\, \mathrm d\lambda \right) \cdot \left(\mathbb E[S] + \mathbb E[T] - 1 + \int_{[0, 1]} F_S \cdot F_T\, \mathrm d\lambda\right). \end{aligned}$

This quantity evaluates the expected closeness that X feels with Y.

(Click for Solution)

Solution. We notice that all random variables have expectation bounded in $[0, 1]$ . Define

$V := \min\{ Q, R \},\quad A := \min\{ S, T \}.$

We first evaluate $\mathbb E[V]$ . By definition,

$\begin{aligned} \mathbb P(V > v) &= \mathbb P(\min\{Q, R\} > v) \\ &= \mathbb P( Q > v, R > v ) \\ &= \mathbb P( Q > v ) \cdot \mathbb P( R > v ). \end{aligned}$

In particular,

$\begin{aligned}(F_Q \cdot F_R)(v) &= F_Q(v) \cdot F_R(v) \\ &= (1 - \mathbb P(Q > v)) \cdot (1 - \mathbb P(R > v)) \\ &= 1 - \mathbb P(Q > v) - \mathbb P(R > v) + \mathbb P(Q > v) \cdot \mathbb P(R > v) \\ &= 1 - \mathbb P(Q > v) - \mathbb P(R > v) + \mathbb P(V > v). \end{aligned}$

Taking integrals on all sides, using the tail-probability characterisation of the expectation, and performing algebruh,

$\displaystyle \mathbb E[V] = \mathbb E[Q] + \mathbb E[R] - 1 + \int_{[0, 1]} (F_Q \cdot F_R)(v)\, \mathrm dv.$

Similarly,

$\displaystyle \mathbb E[A] = \mathbb E[S] + \mathbb E[T] - 1 + \int_0^1 (F_S \cdot F_T)(a)\, \mathrm da.$

Since $Q,R,S,T$ are jointly independent, $V = \min\{Q, R\}$ and $A = \min\{S, T\}$ are also independent, and

$\mathbb E[C] = \mathbb E[V \cdot A] = \mathbb E[V] \cdot \mathbb E[A],$

giving the desired result, after replacing relevant dummy variables.

—Joel Kindiak, 7 Aug 25, 2324H
August 9, 2025