91Ó°ÊÓ

X and Y are the two random variables, and the joint p.d.f. is given as,

\(f\left( {x,y} \right) = \frac{1}{{2\pi }}{e^{ - \left( {1/2} \right)\left( {{x^2} + {y^2}} \right)}}\;\;\;for\; - \infty < x < \infty \;\;and\; - \infty < y < \infty .\)

By the property of Normal distribution, if X and Y are independent standard normal variates, then\[\left( {X + Y} \right)\]they will have a standard normal distribution.

Therefore,

\(E\left( {X + Y} \right) = E\left( X \right) + E\left( Y \right)\)

But, \(E\left( X \right) = 0\)and \(E\left( Y \right) = 0\)

It implies that,

\(E\left( {X + Y} \right) = 0\)

Now,

\(V\left( {X + Y} \right) = V\left( X \right) + V\left( Y \right)\)

But, \(V\left( X \right) = 1\) and \(V\left( Y \right) = 1\)

It implies that,

\(V\left( {X + Y} \right) = 2\)

Let, standard normal variate as,

\[\begin{array}{c}Z = \frac{{\left( {X + Y} \right) - E\left( {X + Y} \right)}}{{\sqrt {V\left( {X + Y} \right)} }}\\ = \frac{{\left( {X + Y} \right) - 0}}{{\sqrt 2 }}\\ = \frac{{\left( {X + Y} \right)}}{{\sqrt 2 }}\end{array}\]

Now,

\(\begin{array}{c}\Pr \left( { - \sqrt 2 < X + Y < 2\sqrt 2 } \right) = \Pr \left( { - 1 < \frac{{X + Y}}{{\sqrt 2 }} < 2} \right)\\ = \Pr \left( { - 1 < Z < 2} \right)\\ = \Pr \left( {Z < 2} \right) - \Pr \left( {Z < - 1} \right)\end{array}\)

\[\begin{array}{c} = \Phi \left( 2 \right) - \left[ {1 - \Phi \left( 1 \right)} \right]\\ = 0.9772 - \left( {1 - 0.8413} \right)\\ = 0.8185\end{array}\]

Therefore,

\(\Pr \left( { - \sqrt 2 < X + Y < 2\sqrt 2 } \right) = 0.8185\;\)