Problem Statement:

We are tasked with proving by induction that if $e_i \sim \text{Bernoulli}(x)$ (for $i = 1, 2, \dots, n$, i.e., they are independent and identically distributed (i.i.d)), then the sum $S_n = \sum_{i=1}^{n} e_i$ is distributed according to a Bernoulli distribution with parameter $p = 0.5 - 0.5(1 - 2x)^n$.

We want to prove: $S_n \sim \text{Bernoulli}(0.5 - 0.5(1-2x)^n)$

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

For $n = 1$, we have $S_1 = e_1$, and by assumption, $e_1 \sim \text{Bernoulli}(x)$. Therefore, the probability that $e_1 = 1$ is $x$, and the probability that $e_1 = 0$ is $1 - x$.

Now, for $n = 1$, the desired result is: $S_1 \sim \text{Bernoulli}(0.5 - 0.5(1 - 2x))$ Let’s check whether this holds.

We can simplify the parameter: $0.5 - 0.5(1 - 2x) = 0.5 - 0.5 + x = x$ So, for $n = 1$, we have $S_1 \sim \text{Bernoulli}(x)$, which matches our assumption. Hence, the base case holds.

Step 2: Induction Hypothesis

Assume that the statement holds for some arbitrary $n$, i.e., assume that: $S_n = \sum_{i=1}^{n} e_i \sim \text{Bernoulli}(0.5 - 0.5(1 - 2x)^n)$

Step 3: Induction Step

We need to show that the statement holds for $n + 1$, i.e., that: $S_{n+1} = \sum_{i=1}^{n+1} e_i \sim \text{Bernoulli}(0.5 - 0.5(1 - 2x)^{n+1})$

We can express $S_{n+1}$ as: $S_{n+1} = S_n + e_{n+1}$ where $S_n$ and $e_{n+1}$ are independent, and $e_{n+1} \sim \text{Bernoulli}(x)$.

Next, we need to compute the distribution of the sum of these two independent random variables $S_n$ and $e_{n+1}$. Since both $S_n$ and $e_{n+1}$ are Bernoulli-distributed, we can use the following result: the sum of two independent Bernoulli random variables, $S_n$ and $e_{n+1}$, is Bernoulli-distributed with some new parameter, say $p'$.

Now, the parameter $p'$ can be computed as follows: $p' = P(S_{n+1} = 1) = P(S_n = 1, e_{n+1} = 0) + P(S_n = 0, e_{n+1} = 1)$

Using the induction hypothesis, we know: $P(S_n = 1) = 0.5 - 0.5(1 - 2x)^n$ and $P(S_n = 0) = 1 - \left(0.5 - 0.5(1 - 2x)^n\right) = 0.5 + 0.5(1 - 2x)^n$

Since $e_{n+1} \sim \text{Bernoulli}(x)$, we also know: $P(e_{n+1} = 1) = x \quad \text{and} \quad P(e_{n+1} = 0) = 1 - x$

Thus, we can compute the probabilities for the two cases: $P(S_n = 1, e_{n+1} = 0) = \left(0.5 - 0.5(1 - 2x)^n\right)(1 - x)$ and $P(S_n = 0, e_{n+1} = 1) = \left(0.5 + 0.5(1 - 2x)^n\right)x$

So, the parameter $p'$ for $S_{n+1}$ becomes: $p' = \left(0.5 - 0.5(1 - 2x)^n\right)(1 - x) + \left(0.5 + 0.5(1 - 2x)^n\right)x$

Simplifying this expression: $p' = 0.5(1 - x) - 0.5(1 - 2x)^n(1 - x) + 0.5x + 0.5(1 - 2x)^n x$ $p' = 0.5 - 0.5(1 - 2x)^n$

Finally, observe that this is the same form as $0.5 - 0.5(1 - 2x)^{n+1}$, since: $1 - 2x = (1 - 2x)$ so: $p' = 0.5 - 0.5(1 - 2x)^{n+1}$

Conclusion

Thus, by induction, we have shown that $S_n \sim \text{Bernoulli}(0.5 - 0.5(1 - 2x)^n)$ holds for all $n$.

Follow-up Questions:

1. How does the parameter of the Bernoulli distribution change as $n$ increases?
2. Why is the sum of two independent Bernoulli random variables still Bernoulli?
3. Can you describe the behavior of the Bernoulli parameter when $x = 0.5$?
4. How would the proof differ if the $e_i$ were not identically distributed?
5. What is the significance of the term $(1 - 2x)$ in the parameter of the Bernoulli distribution?

Tip:

When working with sums of independent Bernoulli variables, leveraging known results about their distributions can greatly simplify the analysis.