Question

Suppose that the joint p.d.f. of two random variablesXandYis

\[f\left( {x,y} \right) = \frac{1}{{2\pi }}{e^{ - \frac{1}{2}\left( {{x^2} + {y^2}} \right)}}; - \infty < x < \infty , - \infty < y < \infty \]

Find\[\Pr \left( { - \sqrt 2 < x + y < 2\sqrt 2 } \right)\].

Short answer

\[P\left( { - \sqrt 2 < X + Y < 2\sqrt 2 } \right) = 0.8186\]

Step-by-step solution

Given information

X and Y are both random variables.

Calculate the probability 

X + Y has a normal distribution.

\[\begin{array}{l}E\left( {X + Y} \right) = 0 + 0 = 0\\V\left( {X + Y} \right) = 1 + 1 = 2\end{array}\]

Z will have standard normal distribution if

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

Then,

\[\begin{array}{c}P\left( { - \sqrt 2 < X + Y < 2\sqrt 2 } \right) = P\left( { - 1 < Z < 2} \right)\\ = P\left( {Z < 2} \right) - P\left( {Z < - 1} \right)\\ = \Phi \left( 2 \right) - \Phi \left( { - 1} \right)\\ = \Phi \left( 2 \right) - 1 + \Phi \left( 1 \right)\\ = 0.8186\end{array}\]

Hence, \[P\left( { - \sqrt 2 < X + Y < 2\sqrt 2 } \right) = 0.8186\].