Question

Suppose that\({X_1}and\,{X_2}\) have the bivariate normal distribution with means\({\mu _1}and\,{\mu _2}\) variances\({\sigma _1}^2and\,{\sigma _2}^2\), and correlation ρ. Determine the distribution of\({X_1} - 3{X_2}\).

Short answer

The random variable \({X_1} - 3{X_2}\)has the normal distribution with mean\({\mu _1} - 3\,{\mu _2}\)and variance\({\sigma _1}^2 + 9{\sigma _2}^2 - 6\rho {\sigma _1}{\sigma _2}\)

Step-by-step solution

Step 1:Given information

\({X_1}{\rm{and}}\,{X_2}\) have the bivariate normal distribution with means\({\mu _1}{\rm{and}}\,{\mu _2}\) variances\({\sigma _1}^2{\rm{and}}\,{\sigma _2}^2\), and correlation ρ. We need to find out the distribution of \({X_1} - 3{X_2}\).

Step-2: Distribution of\({X_1} - 3{X_2}\).

Linear combination of bivariate normal formula: Two random variables \({X_1}{\rm{and}}\,{X_2}\)have a bivariate normal distribution, Let Y =\({a_1}{X_1} + {a_2}{X_2} + b\), where \({a_1},{a_2}\) and b are arbitrary given constants. Then Y has the normal distribution with mean \({a_1}{\mu _1} + {a_2}{\mu _2} + b\) and variance\({a_1}^2{\sigma _1}^2 + {a_2}^2{\sigma _2}^2 + 2{a_1}{a_2}\rho {\sigma _1}{\sigma _2}\)

\({a_1} = 1,{a_2} = - 3,b = 0\), putting these value with the help of the above formula we get that the random variable \({X_1} - 3{X_2}\)has the normal distribution with mean\({\mu _1} - 3\,{\mu _2}\)and variance \({\sigma _1}^2 + 9{\sigma _2}^2 - 6\rho {\sigma _1}{\sigma _2}\)