Question

Find an expression for the mean and variance of y−g(x) for an arbitrary continuous random variable x.

a) Mean=Ey−g(x)=Ey−Eg(x), Variance=Vary−g(x)=Vary+Varg(x)
b) Mean=Ey−g(x)=Ey+Eg(x), Variance=Vary−g(x)=Vary−Varg(x)
c)Mean=Ey−g(x)=Ey⋅Eg(x), Variance=Vary−g(x)=Vary⋅Varg(x)
d) Mean=Ey−g(x)=Ey−Eg(x), Variance=Vary−g(x)=Vary−Varg(x)

Answer 1

The mean of y-g(x) is E[y]-E[g(x)], and the variance is Var[y]+Var[g(x)], given that y and g(x) are independent. Hence, the correct answer is (a).

Step-by-step explanation:

We are looking to find the expression for the mean and variance of y-g(x), where y is a continuous random variable and g(x) is a function of another continuous random variable x. The correct expressions for the mean and variance are based on the properties of expected values and variances.

The mean of y-g(x) is given by:

E[y-g(x)] = E[y] - E[g(x)]

This is because the expected value operator E is linear, meaning that it distributes over addition and subtraction.

The variance of y-g(x) is given by:

Var[y-g(x)] = Var[y] + Var[g(x)]

The variance of a sum (or difference) of two independent random variables is the sum of their variances. This holds true assuming that y and g(x) are independent. If they are dependent, this relationship does not necessarily hold.

Therefore, the correct answer is:

• Mean: E[y-g(x)] = E[y] - E[g(x)]
• Variance: Var[y-g(x)] = Var[y] + Var[g(x)]

So, the answer is a) Mean=E[y-g(x)]=E[y]-E[g(x)], Variance=Var[y-g(x)]=Var[y]+Var[g(x)].