Let e_i ∼ Ber(x) , i = 1, . . . , n be identically, independently distributed. Then we have \sum_{i=1}^{n} e_i ∼ Ber(0.5 - 0.5(1-2x)^n). proof by induction.

### 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.

Let e_i ∼ Ber(x), i = 1, . . . , n be identically, independently distributed. Then we have \sum_{i=1}^{n} e_i ∼ Ber(0.5 - 0.5(1-2x)^n). Proof by induction.

This problem demonstrates the induction proof that the sum of independent, identically distributed Bernoulli random variables follows a Bernoulli distribution with parameter p = 0.5 - 0.5(1 - 2x)^n. Includes detailed step-by-step calculations. Suitable for graduate-level probability theory students.

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