The set given is {(1/x,x=1,2,3,...k),(0,otherwise)}, where k is a positive integer. To find the MGF of Y=2X, we need to first find the PMF of Y.Since Y=2X, the possible values of Y are even integers between 0 and 2k, inclusive. Let y be such an even integer. Then, we have:P(Y=y)=P(2X=y)=P(X=y/2)If y=0, then P(Y=0)=P(X=0)=0 since X takes only positive values. If y=2,4,…,2k, then we have:P(Y=y)=P(X=y/2)={1/kif&nbsp;y=2,4,…,k0otherwiseThus, the PMF of Y is given by:PY(y)={1/kif&nbsp;y=2,4,…,k0otherwiseNow, let's find the MGF of Y. We have:MY(t)=E(etY)=∑y=02ketyPY(y)=e0PY(0)+∑y=2kety1k=1k∑y=0ke2ty=1k(1+e2+e4+⋯+e2kt)This is a finite geometric series with first term 1 and common ratio e2, so we can use the formula for the sum of a finite geometric series to get:MY(t)=1k·1−e2(k+1)t1−e2tTherefore, the MGF of Y is:MY(t)=1−e2(k+1)tk(1−e2t)

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mutheu.stella01

Answered question

2022-05-02

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Answer & Explanation

Don Sumner

Skilled2023-05-04Added 184 answers

The set given is

{(1 / x, x = 1, 2, 3, . . . k), (0, otherwise)}

, where

k

is a positive integer. To find the MGF of

Y = 2 X

, we need to first find the PMF of

Y

.
Since

Y = 2 X

, the possible values of

Y

are even integers between 0 and

2 k

, inclusive. Let

y

be such an even integer. Then, we have:

P (Y = y) = P (2 X = y) = P (X = y / 2)

y = 0

, then

P (Y = 0) = P (X = 0) = 0

since

X

takes only positive values. If

y = 2, 4, \dots, 2 k

, then we have:

P (Y = y) = P (X = y / 2) = {\begin{matrix} 1 / k & if y = 2, 4, \dots, k \\ 0 & otherwise \end{matrix}

Thus, the PMF of

Y

is given by:

P_{Y} (y) = {\begin{matrix} 1 / k & if y = 2, 4, \dots, k \\ 0 & otherwise \end{matrix}

Now, let's find the MGF of

Y

. We have:

M_{Y} (t) = E (e^{t Y}) = \sum_{y = 0}^{2 k} e^{t y} P_{Y} (y)

= e^{0} P_{Y} (0) + \sum_{y = 2}^{k} e^{t y} \frac{1}{k}

= \frac{1}{k} \sum_{y = 0}^{k} e^{2 t y}

= \frac{1}{k} (1 + e^{2} + e^{4} + \dots + e^{2 k t})

This is a finite geometric series with first term 1 and common ratio

e^{2}

, so we can use the formula for the sum of a finite geometric series to get:

M_{Y} (t) = \frac{1}{k} \cdot \frac{1 - e^{2 (k + 1) t}}{1 - e^{2 t}}

Therefore, the MGF of

Y

is:

M_{Y} (t) = \frac{1 - e^{2 (k + 1) t}}{k (1 - e^{2 t})}

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