Step 1: Concept Explanation:
The expected value, E(Y), of a random variable Y is derived from its moment generating function (MGF), \(M_Y(t)\). E(Y) equals the first derivative of the MGF evaluated at t=0. There appears to be an error in the OCR of the MGF. Assuming a standard form such as \( M_Y(t) = \frac{1}{3}e^t + \frac{2}{3}e^{2t} \), or \( M_Y(t) = (\frac{1}{3}e^t + \frac{2}{3})^k \) for some k, let's test a simple discrete distribution. If Y is 1 with probability 1/3 and 2 with probability 2/3, its MGF is \( M_Y(t) = \frac{1}{3}e^{1t} + \frac{2}{3}e^{2t} \). We will calculate E(Y) for this.
Step 2: Formula:
The expected value is the first moment, found by taking the first derivative of the MGF and evaluating at t=0.
\[ E(Y) = M'_Y(t) \Big|_{t=0} \]
Step 3: Detailed Solution:
Assume Y is discrete: it takes the value 1 with probability p and 2 with probability (1-p). The MGF is \( M_Y(t) = p . e^{1t} + (1-p) . e^{2t} \). The answer choices are simple fractions.
Re-examining the OCR'd MGF: \( \frac{1}{3}e^t(e^t - \frac{2}{3}) = \frac{1}{3}e^{2t} - \frac{2}{9}e^t \). This is invalid because \( M_Y(0) = \frac{1}{3} - \frac{2}{9} = \frac{1}{9} eq 1 \).
There must be a typo. Assume a Bernoulli-like structure. A common MGF is \( M_Y(t) = (q + pe^t)^n \). The options suggest a simple answer.
Suppose the question intended a MGF for a variable with two values. If P(Y=1)=1/3 and P(Y=2)=2/3, then \( E(Y) = 1 . \frac{1}{3} + 2 . \frac{2}{3} = \frac{1+4}{3} = \frac{5}{3} \).
Let's try another combination: If \( M_Y(t) = \frac{2}{3} + \frac{1}{3}e^{2t} \), then \( M_Y(0)=1 \). \( E(Y) = \frac{2}{3}e^{2t}|_{t=0} = \frac{2}{3} \). This corresponds to option B.
Let's try \( M_Y(t) = \frac{1}{2} + \frac{1}{2}e^{3t} \). Then \( M_Y(0)=1 \). \( E(Y) = \frac{3}{2}e^{3t}|_{t=0} = \frac{3}{2} \). This corresponds to option D.
Given the ambiguity, we must infer the MGF. The OCR is mathematically incorrect. If we assume the MGF represents a discrete variable with values \(y_1, y_2, \ldots\) and probabilities \(p_1, p_2, \ldots\), such that \(M_Y(t) = \sum p_i e^{y_i t}\). Then, for \(M_Y(0)=1\), \(\sum p_i = 1\). The OCR fails this. If we assume a typo, and that the intended MGF was for a variable Y that is 3/2 times a Bernoulli(1/2) variable, the possibilities are vast. Assume a simple structure that matches an answer. If \(M_Y(t)\) was for Y with \(P(Y=0)=1/2\) and \(P(Y=3)=1/2\), then \(M_Y(t) = \frac{1}{2} + \frac{1}{2}e^{3t}\). Then \(E(Y) = 0 . \frac{1}{2} + 3 . \frac{1}{2} = \frac{3}{2}\).
Let's take the derivative of the OCR'd text anyway and evaluate at 0:\( M_Y(t) = \frac{1}{3}e^{2t} - \frac{2}{9}e^t \)\( M'_Y(t) = \frac{2}{3}e^{2t} - \frac{2}{9}e^t \)\( M'_Y(0) = \frac{2}{3}e^{0} - \frac{2}{9}e^{0} = \frac{2}{3} - \frac{2}{9} = \frac{6-2}{9} = \frac{4}{9} \). This is not in the options.
Due to the clear error in the MGF, there's no correct solution path. In an exam, a common form relates to Bernoulli/Binomial. If we consider a single trial with success (value 1) having probability p and failure (value 0) having probability q, but the values are scaled (Y=aX+b), we get different means. Given the options, 3/2 is plausible. Without a correct MGF, we can't proceed logically.
Let's assume \(M_Y(t) = (\frac{2}{3} + \frac{1}{3}e^t)^2\). This is the MGF of Bin(2, 1/3). The mean would be \(np = 2/3\).Let's assume \(M_Y(t) = (\frac{1}{2} + \frac{1}{2}e^t)^3\). This is the MGF of Bin(3, 1/2). The mean would be \(np = 3/2\). This is a plausible intended question.
Step 4: Answer:
Assuming the question had a typo and intended a Binomial(3, 1/2) distribution, the expected value would be \(np = 3 \times 1/2 = 3/2\).