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How to find the MGF of this pmf?

0

$begingroup$

A random variable $X$ has pmf $p(x;alpha) = (1-alpha)^{x-1} alpha$ for $x = 1,2,dots$. Find the moment generating function of $X$, $M_X(t)$.

What I've done:

$E[e^{tx}] = sum_x e^{tx} (1-alpha)^{x-1} alpha = frac{alpha}{1-alpha} sum_x (e^t (1-alpha))^x$.

Assume $e^t(1-alpha) < 1$. Then the series is geometric with $a_0 = 1$, $r < 1$ hence the sum is

$frac{a_0}{1-r} = frac{1}{(1-alpha)}frac{1}{1-(e^t(1-alpha))}$

But this is wrong. Answer should be $frac{alpha e^t}{1-(1-alpha)e^t}$. Where have i erred?

moment-generating-functions

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asked Mar 25 at 10:24

goblinbgoblinb

796

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add a comment | 

0

$begingroup$

A random variable $X$ has pmf $p(x;alpha) = (1-alpha)^{x-1} alpha$ for $x = 1,2,dots$. Find the moment generating function of $X$, $M_X(t)$.

What I've done:

$E[e^{tx}] = sum_x e^{tx} (1-alpha)^{x-1} alpha = frac{alpha}{1-alpha} sum_x (e^t (1-alpha))^x$.

Assume $e^t(1-alpha) < 1$. Then the series is geometric with $a_0 = 1$, $r < 1$ hence the sum is

$frac{a_0}{1-r} = frac{1}{(1-alpha)}frac{1}{1-(e^t(1-alpha))}$

But this is wrong. Answer should be $frac{alpha e^t}{1-(1-alpha)e^t}$. Where have i erred?

moment-generating-functions

share|cite|improve this question

asked Mar 25 at 10:24

goblinbgoblinb

796

$endgroup$

add a comment | 

$begingroup$

A random variable $X$ has pmf $p(x;alpha) = (1-alpha)^{x-1} alpha$ for $x = 1,2,dots$. Find the moment generating function of $X$, $M_X(t)$.

What I've done:

$E[e^{tx}] = sum_x e^{tx} (1-alpha)^{x-1} alpha = frac{alpha}{1-alpha} sum_x (e^t (1-alpha))^x$.

Assume $e^t(1-alpha) < 1$. Then the series is geometric with $a_0 = 1$, $r < 1$ hence the sum is

$frac{a_0}{1-r} = frac{1}{(1-alpha)}frac{1}{1-(e^t(1-alpha))}$

But this is wrong. Answer should be $frac{alpha e^t}{1-(1-alpha)e^t}$. Where have i erred?

moment-generating-functions

share|cite|improve this question

asked Mar 25 at 10:24

goblinbgoblinb

796

$endgroup$

share|cite|improve this question

asked Mar 25 at 10:24

goblinbgoblinb

796

share|cite|improve this question

asked Mar 25 at 10:24

goblinbgoblinb

796

asked Mar 25 at 10:24

goblinbgoblinb

796

asked Mar 25 at 10:24

goblinbgoblinb

796

2 Answers
2

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1

$begingroup$

Since $x$ starts at $1$, your initial term should be $e^t(1-alpha)$. Then we get the MGF as $$frac{alpha}{1-alpha}frac{e^t(1-alpha)}{1-e^t(1-alpha)}=frac{alpha e^t}{1-(1-alpha)e^t},$$as expected.

share|cite|improve this answer

answered Mar 25 at 10:29

J.G.J.G.

33.7k23252

$endgroup$

add a comment | 

1

$begingroup$

$$sum_{x=1}^{infty}e^{tx}(1-alpha)^{x-1}alpha=alpha e^tsum_{x=0}^{infty}e^{tx}(1-alpha)^x=frac{alpha e^t}{1-e^t(1-alpha)}$$

Your mistake is overlooking that $x$ starts at $1$ (hence not at $0$).

share|cite|improve this answer

answered Mar 25 at 10:35

drhabdrhab

104k545136

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add a comment | 

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1

$begingroup$

Since $x$ starts at $1$, your initial term should be $e^t(1-alpha)$. Then we get the MGF as $$frac{alpha}{1-alpha}frac{e^t(1-alpha)}{1-e^t(1-alpha)}=frac{alpha e^t}{1-(1-alpha)e^t},$$as expected.

share|cite|improve this answer

answered Mar 25 at 10:29

J.G.J.G.

33.7k23252

$endgroup$

add a comment | 

1

$begingroup$

Since $x$ starts at $1$, your initial term should be $e^t(1-alpha)$. Then we get the MGF as $$frac{alpha}{1-alpha}frac{e^t(1-alpha)}{1-e^t(1-alpha)}=frac{alpha e^t}{1-(1-alpha)e^t},$$as expected.

share|cite|improve this answer

answered Mar 25 at 10:29

J.G.J.G.

33.7k23252

$endgroup$

add a comment | 

$begingroup$

Since $x$ starts at $1$, your initial term should be $e^t(1-alpha)$. Then we get the MGF as $$frac{alpha}{1-alpha}frac{e^t(1-alpha)}{1-e^t(1-alpha)}=frac{alpha e^t}{1-(1-alpha)e^t},$$as expected.

share|cite|improve this answer

answered Mar 25 at 10:29

J.G.J.G.

33.7k23252

$endgroup$

share|cite|improve this answer

answered Mar 25 at 10:29

J.G.J.G.

33.7k23252

answered Mar 25 at 10:29

J.G.J.G.

33.7k23252

answered Mar 25 at 10:29

J.G.J.G.

33.7k23252

1

$begingroup$

$$sum_{x=1}^{infty}e^{tx}(1-alpha)^{x-1}alpha=alpha e^tsum_{x=0}^{infty}e^{tx}(1-alpha)^x=frac{alpha e^t}{1-e^t(1-alpha)}$$

Your mistake is overlooking that $x$ starts at $1$ (hence not at $0$).

share|cite|improve this answer

answered Mar 25 at 10:35

drhabdrhab

104k545136

$endgroup$

add a comment | 

1

$begingroup$

$$sum_{x=1}^{infty}e^{tx}(1-alpha)^{x-1}alpha=alpha e^tsum_{x=0}^{infty}e^{tx}(1-alpha)^x=frac{alpha e^t}{1-e^t(1-alpha)}$$

Your mistake is overlooking that $x$ starts at $1$ (hence not at $0$).

share|cite|improve this answer

answered Mar 25 at 10:35

drhabdrhab

104k545136

$endgroup$

add a comment | 

$begingroup$

$$sum_{x=1}^{infty}e^{tx}(1-alpha)^{x-1}alpha=alpha e^tsum_{x=0}^{infty}e^{tx}(1-alpha)^x=frac{alpha e^t}{1-e^t(1-alpha)}$$

Your mistake is overlooking that $x$ starts at $1$ (hence not at $0$).

share|cite|improve this answer

answered Mar 25 at 10:35

drhabdrhab

104k545136

$endgroup$

share|cite|improve this answer

answered Mar 25 at 10:35

drhabdrhab

104k545136

answered Mar 25 at 10:35

drhabdrhab

104k545136

answered Mar 25 at 10:35

drhabdrhab

104k545136