To prove the probability mass function (PMF) of a Binomial distribution using induction and convolutions on an i.i.d. sequence of random variables $X_1, \dots, X_n \sim \text{Bernoulli}(p)$, we will utilize the following steps and Pascal's combinatorial identity.

Step 1: Definition of Bernoulli and Binomial Distributions

Let $X_1, X_2, \dots, X_n$ be i.i.d. Bernoulli random variables, where each $X_i$ takes the value 1 with probability $p$ and 0 with probability $1 - p$. That is, for each $i$: $P(X_i = 1) = p \quad \text{and} \quad P(X_i = 0) = 1 - p.$ The sum of these i.i.d. Bernoulli random variables is a Binomial random variable: $S_n = X_1 + X_2 + \dots + X_n.$ The random variable $S_n$ follows a Binomial distribution, denoted $S_n \sim \text{Binomial}(n, p)$, with PMF: $P(S_n = k) = \binom{n}{k} p^k (1 - p)^{n - k}, \quad k = 0, 1, \dots, n.$

Step 2: Base Case (Induction for $n = 1$)

For $n = 1$, we know that $S_1 = X_1$, and since $X_1 \sim \text{Bernoulli}(p)$, its PMF is given by: $P(S_1 = 0) = P(X_1 = 0) = 1 - p,$ $P(S_1 = 1) = P(X_1 = 1) = p.$ This corresponds to the Binomial PMF for $n = 1$: $P(S_1 = k) = \binom{1}{k} p^k (1 - p)^{1 - k}, \quad k = 0, 1.$ Thus, the base case holds.

Step 3: Induction Hypothesis

Assume that for some $n \geq 1$, the PMF of $S_n$ is: $P(S_n = k) = \binom{n}{k} p^k (1 - p)^{n - k}, \quad k = 0, 1, \dots, n.$ We will prove that the PMF of $S_{n+1} = S_n + X_{n+1}$ also follows a Binomial distribution.

Step 4: Inductive Step (Using Convolutions)

We express $S_{n+1} = S_n + X_{n+1}$. Since $X_{n+1} \sim \text{Bernoulli}(p)$, the PMF of $S_{n+1}$ is the convolution of the PMFs of $S_n$ and $X_{n+1}$. The convolution formula is: $P(S_{n+1} = k) = P(S_n + X_{n+1} = k) = P(S_n = k) P(X_{n+1} = 0) + P(S_n = k-1) P(X_{n+1} = 1).$ Substituting the Bernoulli probabilities: $P(S_{n+1} = k) = P(S_n = k) (1 - p) + P(S_n = k - 1) p.$ Using the induction hypothesis, we know: $P(S_n = k) = \binom{n}{k} p^k (1 - p)^{n - k} \quad \text{and} \quad P(S_n = k - 1) = \binom{n}{k-1} p^{k-1} (1 - p)^{n - (k - 1)}.$ Thus, we can write: $P(S_{n+1} = k) = \binom{n}{k} p^k (1 - p)^{n - k + 1} + \binom{n}{k-1} p^k (1 - p)^{n - k + 1}.$ Factoring out the common terms: $P(S_{n+1} = k) = p^k (1 - p)^{n - k + 1} \left( \binom{n}{k} + \binom{n}{k-1} \right).$ Using Pascal's identity $\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$, we get: $P(S_{n+1} = k) = \binom{n+1}{k} p^k (1 - p)^{(n + 1) - k}.$ Thus, the PMF of $S_{n+1}$ matches the Binomial PMF with parameters $n+1$ and $p$.

Step 5: Conclusion

By the principle of mathematical induction, the PMF of the sum of $n$ i.i.d. Bernoulli random variables, $S_n = X_1 + X_2 + \dots + X_n$, follows the Binomial distribution: $P(S_n = k) = \binom{n}{k} p^k (1 - p)^{n - k}, \quad k = 0, 1, \dots, n.$

This completes the proof.

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Would you like more details on any step? Here are some related questions you can explore:

1. How does Pascal's triangle identity help in the convolution of Bernoulli variables?
2. What is the significance of convolution in probability theory?
3. Can the proof be generalized for other distributions using convolution?
4. How would the proof change for non-i.i.d. Bernoulli variables?
5. What are other ways to prove the PMF of a Binomial distribution?

Tip: Always pay attention to the base case and how the induction hypothesis is used in the inductive step, as this ensures the proof works for all $n$.