Final answer:
To find E[x], differentiate the moment generating function and evaluate it at t=0. The answer is 0. To find Var[x], differentiate the moment generating function twice and evaluate it at t=0. The answer is 4. To find the moment generating function my(t) of y = 3x + 4 for t=3, substitute t=3 into the given mx(t) function.
Step-by-step explanation:
To find the answer to part (a), we can use the definition of the moment generating function. The moment generating function is defined as mx(t) = E[e^(tx)], where E is the expectation operator, and x is the random variable. By differentiating mx(t) with respect to t and evaluating it at t=0, we can find E[x]. In this case, mx(t) = (1/2)e^t + (1/3)e^(-3t) + (1/6)e^(3t).
(a) Find E[x]:
By differentiating mx(t) with respect to t:
mx'(t) = (1/2)e^t + (1/3)(-3)e^(-3t) + (1/6)(3)e^(3t).
Evaluating at t=0:
mx'(0) = (1/2) + (1/3)(-3) + (1/6)(3) = 1/2 - 1 + 1/2 = 0.
Therefore, E[x] = mx'(0) = 0.
To find the answer to part (b), we can use the definition of the variance. The variance is defined as Var[x] = E[(x-E[x])^2]. By differentiating the moment generating function twice and evaluating it at t=0, we can find Var[x].
(b) Find Var[x]:
By differentiating mx(t) with respect to t twice:
mx''(t) = (1/2)e^t + (1/3)(-3)(-3)e^(-3t) + (1/6)(3)(3)e^(3t).
Evaluating at t=0:
mx''(0) = (1/2) + (1/3)(-3)(-3) + (1/6)(3)(3) = 1/2 + 3 + 1/2 = 4.
Therefore, Var[x] = mx''(0) = 4.
To find the answer to part (c), we can use the property of the moment generating function. If Y is a linear transformation of X, then the moment generating function of Y is given by my(t) = m(aX+b), where a and b are constants. In this case, Y = 3X+4, so my(t) = m(3X+4).
(c) Find the moment generating function my(t) of y = 3x + 4 for t = 3:
Substituting t=3 and mx(t) = (1/2)e^t + (1/3)e^(-3t) + (1/6)e^(3t):
my(t) = m(3X+4) = (1/2)e^(3(3)+4) + (1/3)e^(-3(3)+4) + (1/6)e^(3(3)+4) = (1/2)e^13 + (1/3)e^(-5) + (1/6)e^13.