Let X ∼ N(0, 2) and Y ∼ N(0, 1), and let Cov(X, Y ) = −1. What is the distribution of 2X − 3Y ?

Answer 1

Linear combinations of normally distributed random variables (which is what this is) is again a normally distributed random variable. So at a first level of description, we already know the answer is some normal distribution.

We next need to decide the mean and variance of that normal distribution.

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The mean obeys a simple linearity property. That is to say, if we take the means to be µ₁ and µ₂, then the linearity property tells us that the mean of the linear combination is 2µ₁ - 3µ₂. (That is to say, we just use the same coefficients on the means, which are used on the random variables.) In this case, because µ₁ = 0 = µ₂ it follows that the mean of the new distribution is 2*0 - 3*0 = 0.

So the mean is 0.

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The variance satisfies a slightly different condition, which in effect takes squares rather before taking sums. So in this case the variance can be computed by

V(2X - 3Y) = V(2X) + V(-3Y)

= 2² * V(X) + (-3)² * V(Y)

= 4V(X) + 9V(Y)

= 4*2 + 9*1

= 17

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Therefore the answer is that this random variable is distributed as N(0,17).