I have two questions related to moment generating functions. I can't find any sort of examples or explanations in my book...
1) Calculate P(X<=2) if the moment generating function is given by:
Mx(t)=((1/4)+(3/4)e^t)^5
2) Suppose the moment-generating function for a random variable W is given by:
Mw(t)=(e^(-3+3e^t))*((2/3)+(1/3)e^t)^4
Calculate P(W<=1).
To calculate the probability P(X <= 2) or P(W <= 1), we need to first find the probability distribution function (PDF) or cumulative distribution function (CDF) of the random variable. However, in the given questions, we are provided with the moment generating functions (MGFs) of X and W. So, we need to use the MGFs to find the PDF/CDF and then calculate the probabilities.
Let's start by finding the PDF or CDF using the MGFs given in the questions:
1) Calculate P(X <= 2):
We are given the MGF of X as Mx(t) = ((1/4) + (3/4)e^t)^5.
To find the CDF, we need to take the derivative of the MGF with respect to t. The order of the derivative will be the power of the MGF.
So, let's find the fifth derivative of Mx(t):
Mx(t) = ((1/4) + (3/4)e^t)^5
d^5(Mx(t))/dt^5 = (5!) * ((3/4)e^t)^0 * (1/4) * (3/4)e^t
Simplifying this expression, we get:
d^5(Mx(t))/dt^5 = (5!) * (1/4) * (3/4)e^t
= 120 * (3/4)e^t
Now that we have the fifth derivative, the CDF is obtained by evaluating the MGF at the point t = 0 and taking the exponential of the result:
CDF(x) = e^(d^5(Mx(t))/dt^5 * t) evaluated at t = 0
CDF(x) = e^(120 * (3/4)e^t) evaluated at t = 0
CDF(x) = e^(120 * (3/4))
CDF(x) = e^90
Finally, to calculate P(X <= 2), we need to find the probability at x = 2 in the CDF:
P(X <= 2) = CDF(2)
2) Calculate P(W <= 1):
We are given the MGF of W as Mw(t) = (e^(-3+3e^t)) * ((2/3) + (1/3)e^t)^4.
To find the CDF, we need to take the derivative of the MGF with respect to t. The order of the derivative will be the power of the MGF.
So, let's find the fourth derivative of Mw(t):
Mw(t) = (e^(-3+3e^t)) * ((2/3) + (1/3)e^t)^4
d^4(Mw(t))/dt^4 = (4!) * (e^(-3+3e^t)) * ((2/3) + (1/3)e^t)^0 * (1/3) * (1/3)e^t
Simplifying this expression, we get:
d^4(Mw(t))/dt^4 = (4!) * (1/9)e^t
= 24 * (1/9)e^t
Now that we have the fourth derivative, the CDF is obtained by evaluating the MGF at the point t = 0 and taking the exponential of the result:
CDF(w) = e^(d^4(Mw(t))/dt^4 * t) evaluated at t = 0
CDF(w) = e^(24 * (1/9)e^t) evaluated at t = 0
CDF(w) = e^(24 * (1/9))
CDF(w) = e^(8/3)
Finally, to calculate P(W <= 1), we need to find the probability at w = 1 in the CDF:
P(W <= 1) = CDF(1)
Note: The above calculations assume that the MGFs provided are valid and represent valid probability distributions. It is always important to check the validity of the MGFs and their corresponding PDFs/CDFs before performing calculations.