Question

Suppose that the random variables \({X_1}\,\,\,{\rm{and}}\,\,\,{X_2}\) are independent and that each has the normal distribution with mean 0 and variance \({\sigma ^2}\) . Determine the value of

\(P\left( {\frac{{{{\left( {{X_1} + {X_2}} \right)}^2}}}{{{{\left( {{X_1} - {X_2}} \right)}^2}}} < 4} \right)\)

Short answer

The of the value of the given probability is 0.7

Step-by-step solution

Given information

that the random variables \({X_1}\,\,\,{\rm{and}}\,\,\,{X_2}\) are independent and that each has the normal distribution with mean 0 and variance \({\sigma ^2}\).It is needed to compute the following

\(P\left( {\frac{{{{\left( {{X_1} + {X_2}} \right)}^2}}}{{{{\left( {{X_1} - {X_2}} \right)}^2}}} < 4} \right)\)

Computation of \(P\left( {\frac{{{{\left( {{X_1} + {X_2}} \right)}^2}}}{{{{\left( {{X_1} - {X_2}} \right)}^2}}} < 4} \right)\) 

\({\left( {{X_1} - {X_2}} \right)^2} = 2\left( {{{\left( {{X_1} - \frac{{{X_1} + {X_2}}}{2}} \right)}^2} + {{\left( {{X_2} - \frac{{{X_1} + {X_2}}}{2}} \right)}^2}} \right)\) and also

\(\begin{align}\frac{{{{\left( {{X_1} + {X_2}} \right)}^2}}}{{{{\left( {{X_1} - {X_2}} \right)}^2}}} < 4\\ \Rightarrow \left| {\frac{{{X_1} + {X_2}}}{{{X_1} - {X_2}}}} \right| < 2\\ \Rightarrow - 2 < \sqrt Y < 2\,\,\,\,\,\,{\rm{where}}\,\,Y = \frac{{{{\left( {{X_1} + {X_2}} \right)}^2}}}{{{{\left( {{X_1} - {X_2}} \right)}^2}}}\end{align}\)

Now

\(\begin{align}P\left( {\frac{{{{\left( {{X_1} + {X_2}} \right)}^2}}}{{{{\left( {{X_1} - {X_2}} \right)}^2}}} < 4} \right)\\ &= P\left( { - 2 < \sqrt Y < 2} \right)\\ &= 0.7\end{align}\)

Here Y follows t-distribution with 2-1=1 degree of freedom and the value of the probability is computed from the table at the end of book.