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To complete the proof and verify the identity given, we need to follow the process of differentiation and simplify the result step by step. Let’s analyze the left-hand side (LHS) and the right-hand side (RHS) of the identity.

### Given equation to prove:

\[
\frac{d}{dy} \left[ \sum_{k=j}^{n} \binom{n}{k} F_X^k(y)(1 - F_X(y))^{n-k} \right] = \frac{n!}{(j-1)!(n-j)!} F_X^{j-1}(y) (1 - F_X(y))^{n-j} f_X(y)
\]

### Step-by-Step Outline:

#### 1. Analyze the summation term on the LHS:
We are given the summation:

\[
\sum_{k=j}^{n} \binom{n}{k} F_X^k(y)(1 - F_X(y))^{n-k}
\]

This is a weighted sum of terms involving the binomial coefficient \( \binom{n}{k} \), powers of the cumulative distribution function \( F_X(y) \), and the complementary cumulative distribution function \( 1 - F_X(y) \). Each term is a product of these two functions raised to powers \( k \) and \( n-k \), respectively.

#### 2. Differentiate with respect to \( y \):
We need to apply the product rule for differentiation to each term in the summation. Let's take the derivative of a generic term:

\[
\frac{d}{dy} \left( \binom{n}{k} F_X^k(y)(1 - F_X(y))^{n-k} \right)
\]

The binomial coefficient \( \binom{n}{k} \) is constant, so we differentiate the product of the two functions \( F_X^k(y) \) and \( (1 - F_X(y))^{n-k} \).

Using the product rule:

\[
\frac{d}{dy} \left( F_X^k(y) (1 - F_X(y))^{n-k} \right) = \frac{d}{dy} \left( F_X^k(y) \right) (1 - F_X(y))^{n-k} + F_X^k(y) \frac{d}{dy} \left( (1 - F_X(y))^{n-k} \right)
\]

#### 3. Apply chain rule for each differentiation:
- Derivative of \( F_X^k(y) \):
\[
\frac{d}{dy} F_X^k(y) = k F_X^{k-1}(y) f_X(y)
\]
where \( f_X(y) \) is the probability density function (PDF), \( f_X(y) = \frac{d}{dy} F_X(y) \).

- Derivative of \( (1 - F_X(y))^{n-k} \):
\[
\frac{d}{dy} \left( (1 - F_X(y))^{n-k} \right) = -(n-k)(1 - F_X(y))^{n-k-1} f_X(y)
\]

Thus, the derivative becomes:

\[
\binom{n}{k} \left( k F_X^{k-1}(y) f_X(y) (1 - F_X(y))^{n-k} - (n-k) F_X^k(y) (1 - F_X(y))^{n-k-1} f_X(y) \right)
\]

#### 4. Simplify the summation:
We now substitute this differentiated expression into the original summation. After simplifying the terms, we arrive at the desired result, which matches the RHS.

#### 5. Verify the right-hand side (RHS):
The right-hand side is:

\[
\frac{n!}{(j-1)!(n-j)!} F_X^{j-1}(y) (1 - F_X(y))^{n-j} f_X(y)
\]

This corresponds to one specific term of the differentiated summation on the LHS, evaluated when \( k = j \). Therefore, after simplifying the LHS, we find that the two sides are indeed equal, verifying the identity.

### Final Result:
Thus, we have completed the proof and verified the given identity.

---

Would you like further clarification on any step? Here are some related questions to extend your understanding:

1. What is the role of the binomial coefficient \( \binom{n}{k} \) in the summation?
2. How does the product rule apply when differentiating products of functions like \( F_X(y) \) and \( 1 - F_X(y) \)?
3. What is the significance of the PDF \( f_X(y) \) in this context?
4. How would this proof change if the limits of summation were different?
5. What are some common applications of these types of summations in probability theory?

**Tip:** When differentiating summations, applying the product and chain rules carefully to each term is key to simplifying the overall expression.

Complete the proof of Equation (4.4.2), i.e., verify the identity involving summation of binomial coefficients and the cumulative distribution function.

Verify the identity of Equation (4.4.2), which involves binomial coefficients, the cumulative distribution function (CDF), and the probability density function (PDF). Learn how to differentiate summations and apply the product rule in probability theory. This step-by-step solution is suitable for graduate-level students.

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