Question

Let m(t)=(1/6)et + (2/6)e2t + (3/6)e3t. Find the following:

a) E(Y)
b) V(Y)
c) The distribution of Y.

Answer 1

a) E(Y) = 0

b) V(Y) = 10

c) The distribution of Y is not normal as it does not have finite variance. However, we can find the probability density function (PDF) of Y using the inverse Laplace transform of m(s):

Y ~ PDF(y) = (1/6)e^{-y/6} + (2/6)e^{-2y/6} + (3/6)e^{-3y/6}

Step-by-step explanation:

To find the expected value (E(Y)) and variance (V(Y)) of a random variable Y, we need to find the moments of Y. In this case, m(t) is a linear combination of three exponential functions, so we can use the moment generating function (MGF) to find the moments. The MGF of m(t) is:

MGF(s) = (1/6)e^{(s/6)} + (2/6)e^{(2s/6)} + (3/6)e^{(3s/6)}

a) To find E(Y), we take the derivative of MGF(s) with respect to s and set s=0:

E(Y) = MGF'(0) = 0

This result may seem counterintuitive since the sum of three exponential functions is not equal to zero. However, this is because the coefficients in front of each exponential function add up to zero.

b) To find V(Y), we take the second derivative of MGF(s) with respect to s and set s=0:

V(Y) = MGF''(0) = 10

c) The variance is finite, but it is not equal to zero, which means that Y is not normally distributed. To find the PDF of Y, we can use the inverse Laplace transform of MGF(s):

Y ~ PDF(y) = (1/6)*e^{-y/6} + (2/6)*e^{-2y/6} + (3/6)*e^{-3y/6}

This PDF is a mixture of three exponential distributions with different rates. The probability density function shows that Y has a long right tail, which means that there is a higher probability of observing large values for Y compared to a normal distribution with the same variance. This non-normality arises due to the presence of multiple exponential components in m(t).

Answer 2

a) The expected value
`\(E(Y)\)` of the random variable (Y) is equal to the mean of the function (m(t)), which is
`\((13)/(6)\)` when simplified.

b) The variance (V(Y)) of the random variable (Y) is found by subtracting the square of the mean from the mean of the squares, resulting in
`\((185)/(36)\).`

c) The distribution of (Y) follows a continuous probability distribution, and to specify it further, additional information or constraints on (t) would be needed.

Step-by-step explanation:

a) To find (E(Y)), the expected value of the random variable (Y), we evaluate the mean of the function (m(t)) by integrating (m(t)) with respect to (t) over its entire range. The expression for (m(t)) is
`\((1)/(6)e^t + (2)/(6)e^(2t) + (3)/(6)e^(3t)\)`. Integrating this function yields
`\((13)/(6)\),` which is the expected value (E(Y)).

b) To find (V(Y)), the variance of (Y), we calculate the mean of the squares of (m(t)) and subtract the square of the mean. The variance formula is
`\(V(Y) = E(Y^2) - [E(Y)]^2\).` After evaluating the necessary integrals, we obtain
`\(V(Y) = (185)/(36)\).`

c) The distribution of (Y) is not explicitly specified in the given information. To describe the distribution more thoroughly, additional constraints or information about the variable (t) would be required. As it stands, we know the expected value and variance, but the distribution type (e.g., normal, exponential) cannot be precisely determined without further details.