Question

Linear operations on jointly Gaussian random variables yield Gaussian random variables: Suppose that the random variables X and Y are jointly Gaussian with E[X]=1,Var(X)=16,E[Y]=−1, Var(Y)=9, and rho X,Y =1/3.

(a) Suppose W=X+5. Find f W (w), the probability density function (pdf) of W.

Answer 1

To find the pdf of W, we use the properties of Gaussian random variables. Since W = X + 5 and X is Gaussian with mean 1 and variance 16, W is also Gaussian with mean 6 (1 + 5) and variance 16. The pdf of W is W~N(6, 16) and expressed as f_W(w) = (1/(4sqrt(2pi)))e^(-(1/32)(w-6)^2).

Step-by-step explanation:

Finding the Probability Density Function of W

To find the probability density function (pdf) of W, where W is defined as W = X + 5, we'll use the properties of Gaussian random variables. Given that X is a Gaussian random variable with E[X] = 1 (mean) and Var(X) = 16 (variance), the transformation W = X + 5 will also be normally distributed as this is a linear transformation.

The mean of W, E[W], will be the mean of X plus 5, which is 1 + 5 = 6. The variance of W will be the same as the variance of X since adding a constant does not change the variance. So, Var(W) = Var(X) = 16. Thus, the distribution of W will be W~N(6, 16).

The pdf of W can be expressed as:

fW(w) = \(\frac{1}{4\sqrt{2\pi}}e^{-\frac{1}{32}(w-6)^2}\), where \(w\) is a real number.