Question

Question: Show that if X₁ , X_2, X_3….. X_n are mutually independent random variables, then\(E\left( {\coprod\limits_{i = 1}^n {{X_i}} } \right) = \coprod\limits_{i = 1}^n {} E({X_i})\).

Short answer

Answer:

The equality\(E\left( {\coprod\limits_{i = 1}^n {{X_i}} } \right) = \coprod\limits_{i = 1}^n {} E({X_i})\)has been proved.

Step-by-step solution

Note the given data

Given X₁ , X_2, X_3….. X_n are mutually independent random variables.

Definition  

The random variables X and Y on a sample space S are independent if\(p\left( {X = {r_1}\begin{array}{*{20}{c}}{}&{andY = {r_2}}\end{array}} \right) = p\left( {X = {r_1}} \right).p\left( {Y = {r_2}} \right)\)

Proof

Since X₁ , X_2, X_3….. X_n are mutually independent random variables.

\(P\left( {\bigcap\limits_{i = 1}^n {{X_i}} = {x_i}} \right) = \prod\limits_{i = 1}^n {P\left( {{X_i} = {x_i}} \right)} \)

\(\begin{array}{l}E\left( {\coprod\limits_{i = 1}^n {{X_i}} } \right) = \sum {{x_1}} {x_2}....{x_n}.P\left( {\bigcap\limits_{i = 1}^n {{X_i}} = {x_i}} \right)\\\begin{array}{*{20}{c}}{}&{}&{}&{}\end{array} = \sum {.\sum {.....} } \sum {{x_1}} {x_2}....{x_n}.\left( {\prod\limits_{i = 1}^n {P\left( {{X_i} = {x_i}} \right)} } \right)\\\begin{array}{*{20}{c}}{}&{}&{}&{}\end{array} = \sum {{x_1}} .P\left( {{X_1} = {x_1}} \right).\sum {{x_2}} .P\left( {{X_2} = {x_2}} \right)......\sum {{x_n}} .P\left( {{X_n} = {x_n}} \right)\\\begin{array}{*{20}{c}}{}&{}&{}&{}\end{array} = \left( {\prod\limits_{i = 1}^n {E\left( {{X_i}} \right)} } \right)\end{array}\)

Conclusion  

Thus,\(E\left( {\coprod\limits_{i = 1}^n {{X_i}} } \right) = \coprod\limits_{i = 1}^n {} E({X_i})\)has been proved.