Question

Suppose that six random variables\({X_1},{X_2},...,{X_6}\)form a random sample from the standard normal distribution, and let

\(Y = {\left( {{X_1} + {X_2} + {X_3}} \right)^2} + {\left( {{X_4} + {X_5} + {X_6}} \right)^2}\). Determine a value ofcsuch that the random variablecYwill have a\({\chi ^2}\)distribution.

Short answer

\(c = \frac{1}{3}\)

Step-by-step solution

Given information

Let\({X_1},{X_2},...,{X_6}\)be a random sample from a standard normal distribution.

Calculate the value of c 

\({X_1} + {X_2} + {X_3}\)And \({X_4} + {X_5} + {X_6}\) both will follow \(N\left( {0,3} \right)\).

\(\frac{{{X_1} + {X_2} + {X_3}}}{{\sqrt 3 }}\)and \(\frac{{{X_4} + {X_5} + {X_6}}}{{\sqrt 3 }}\)will follows a standard normal distribution.

So, each \(\frac{{{{\left( {{X_1} + {X_2} + {X_3}} \right)}^2}}}{3}\) and \(\frac{{{{\left( {{X_4} + {X_5} + {X_6}} \right)}^2}}}{3}\)will follow \({\chi ^2}\)with d.f 1.

And \(\frac{Y}{3} = \frac{{{{\left( {{X_1} + {X_2} + {X_3}} \right)}^2}}}{3} + \frac{{{{\left( {{X_4} + {X_5} + {X_6}} \right)}^2}}}{3}\) will follows \({\chi ^2}\) with 2 degrees of freedom.

Therefore,\(c = \frac{1}{3}\).