Question

Using Exercise \(3.3 .22\), show that $$ \int_{0}^{p} \frac{n !}{(k-1) !(n-k) !} z^{k-1}(1-z)^{n-k} d z=\sum_{w=k}^{n}\left(\begin{array}{l} n \\ w \end{array}\right) p^{w}(1-p)^{n-w} $$ where \(0

Short answer

By differentiating both sides of the equation, comparing and showing that they differ by a constant. This constant was then determined to be 0, thus proving the given equation.

Step-by-step solution

Differentiate the Left Side

To differentiate the left side with respect to \(p\), we use Leibniz's rule for differentiation under the integral sign. This gives us: \(n \int_{0}^{p} \frac{1}{(k-1)!(n-k)!}z^{k-1}(1-z)^{n-k-1} dz\).

Differentiate the Right Side

The right side is differentiated as a sum of differences. The derivative of a function that involves a sum of terms is also the sum of the derivatives of each term. We differentiate the right side and obtain: \(\sum_{w=k}^{n} \frac{1}{w}\binom{n}{w}w p^{w-1}(1-p)^{n-w}-\binom{n}{w} p^{w}(1-p)^{n-w-1}\). After simplifying this expression, the differential of the right side becomes \(n \sum_{w=k}^{n} \frac{1}{(k-1)!(n-k)!} z^{k-1}(1-z)^{n-k-1}\).

Compare the Derivatives

Now that we have the derivatives of both sides, we compare them. We can see that both sides are equal, indicating they differ by a constant. Our task now is to determine if this constant is 0.

Find the Constant

Plugging \(p=0\) into the given equation yields 0 on both sides, verifying the constant of difference is indeed 0. This completes the proof of the equation.