Q.6.27 - A First Course in Probability

Question

Establish Equation $(6.2)$ by differentiating Equation $\left(6.4\right)$.

Step-by-Step Solution

Answer

On differentiating equation $(6.4)$ we get equation $(6.2)$.

Step 1 : Differentiation

Differentiation is a technique for calculating a derivative, which is the rate of change of a function's output $y$ in relation to the change of a variable $x$.

Step 2 : Explanation :

$F_{x_{ij}}\left(y\right)=\sum _{k=j}^n\left(\begin{array}{c}n\\ k\end{array}\right){\left[F\left(y\right)\right]}^k{\left[1-F\left(y\right)\right]}^{n-k}..........(6.4)$

Differentiating both sides with respect to y, then

$$\begin{gathered} f_{X_j}\left(y\right)=\sum _{k=j}^n\left(\begin{array}{l}n\\ k\end{array}\right)kF{\left(y\right)}^{k-1}f\left(y\right)[1-F(y){]}^{n-k}-\sum _{k=j}^n\left(\begin{array}{l}n\\ k\end{array}\right)F{\left(y\right)}^{k-1}(n-k\left)\right[1-F\left(y\right){]}^{n-k-1}f\left(y\right) \\ =\sum _{i=j}^n\frac{n!}{(n-k)!(k-1)!}F{\left(y\right)}^{k-1}f\left(y\right)[1-F(y){]}^{n-k}-\sum _{k=j+1}^n\frac{n!}{(n-k)!(k-1)!}{F}^{k-1}(y\left)f\right(y\left)\right[1-F\left(y\right){]}^{n-k}\text{ by j}=k+1 \\ =\frac{n!}{(n-j)!(j-1)!}{F}^{ j-1}\left(y\right)f\left(y\right)[1-F(y){]}^{n-j} \end{gathered}$$

which is the same as the equation $(6.2)$

Hence proved.