Finding moment generating function from a given probability mass function












0












$begingroup$


Let $Y_1$ and $Y_2$ be two independent discrete random variables such that $p_1(y_1) = frac13$; $y_1 = -2, -1, 0$ and $p_2(y_2) = frac12$, $y_2=1,6$. Let K = $Y_1 + Y_2$. Find the moment generating function of $Y_1,Y_2,$ and $K$.



Attempt:



I know that the moment generating function is just a summation since it is discrete: $$M_x(t) = sum_{-2}^0 e^{ty_1}p_1(y_1)$$



Which then, for the moment generating function of $Y_1$, should be $(frac{e^t}{3})^{-2}+(frac{e^t}{3})^{-1}+(frac{e^t}{3})^{0}$. However I don't think this is the right way to solve the question, but I don't know what I am missing.



The moment generating function of $Y_2$ can be solved using the same method of $Y_1$, but am I right in saying that the moment generating function of K will just be the sum of the moment generating functions of $Y_1$ and $Y_2$?






statistics moment-generating-functions







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$endgroup$












  • $begingroup$
    What is a mgf-pmf?
    $endgroup$
    – Dietrich Burde
    Jan 25 at 20:37










  • $begingroup$
    mgf as in moment generating functions, and pmf as in probability mass function. sorry for the confusion
    $endgroup$
    – peco
    Jan 25 at 20:38










  • $begingroup$
    It should be $e^{-2t}/3+cdots$, not $(e^t/3)^{-2}+cdots$.
    $endgroup$
    – J.G.
    Jan 25 at 23:13
















0












$begingroup$


Let $Y_1$ and $Y_2$ be two independent discrete random variables such that $p_1(y_1) = frac13$; $y_1 = -2, -1, 0$ and $p_2(y_2) = frac12$, $y_2=1,6$. Let K = $Y_1 + Y_2$. Find the moment generating function of $Y_1,Y_2,$ and $K$.



Attempt:



I know that the moment generating function is just a summation since it is discrete: $$M_x(t) = sum_{-2}^0 e^{ty_1}p_1(y_1)$$



Which then, for the moment generating function of $Y_1$, should be $(frac{e^t}{3})^{-2}+(frac{e^t}{3})^{-1}+(frac{e^t}{3})^{0}$. However I don't think this is the right way to solve the question, but I don't know what I am missing.



The moment generating function of $Y_2$ can be solved using the same method of $Y_1$, but am I right in saying that the moment generating function of K will just be the sum of the moment generating functions of $Y_1$ and $Y_2$?






statistics moment-generating-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is a mgf-pmf?
    $endgroup$
    – Dietrich Burde
    Jan 25 at 20:37










  • $begingroup$
    mgf as in moment generating functions, and pmf as in probability mass function. sorry for the confusion
    $endgroup$
    – peco
    Jan 25 at 20:38










  • $begingroup$
    It should be $e^{-2t}/3+cdots$, not $(e^t/3)^{-2}+cdots$.
    $endgroup$
    – J.G.
    Jan 25 at 23:13














0












0








0





$begingroup$


Let $Y_1$ and $Y_2$ be two independent discrete random variables such that $p_1(y_1) = frac13$; $y_1 = -2, -1, 0$ and $p_2(y_2) = frac12$, $y_2=1,6$. Let K = $Y_1 + Y_2$. Find the moment generating function of $Y_1,Y_2,$ and $K$.



Attempt:



I know that the moment generating function is just a summation since it is discrete: $$M_x(t) = sum_{-2}^0 e^{ty_1}p_1(y_1)$$



Which then, for the moment generating function of $Y_1$, should be $(frac{e^t}{3})^{-2}+(frac{e^t}{3})^{-1}+(frac{e^t}{3})^{0}$. However I don't think this is the right way to solve the question, but I don't know what I am missing.



The moment generating function of $Y_2$ can be solved using the same method of $Y_1$, but am I right in saying that the moment generating function of K will just be the sum of the moment generating functions of $Y_1$ and $Y_2$?






statistics moment-generating-functions







share|cite|improve this question











$endgroup$




Let $Y_1$ and $Y_2$ be two independent discrete random variables such that $p_1(y_1) = frac13$; $y_1 = -2, -1, 0$ and $p_2(y_2) = frac12$, $y_2=1,6$. Let K = $Y_1 + Y_2$. Find the moment generating function of $Y_1,Y_2,$ and $K$.



Attempt:



I know that the moment generating function is just a summation since it is discrete: $$M_x(t) = sum_{-2}^0 e^{ty_1}p_1(y_1)$$



Which then, for the moment generating function of $Y_1$, should be $(frac{e^t}{3})^{-2}+(frac{e^t}{3})^{-1}+(frac{e^t}{3})^{0}$. However I don't think this is the right way to solve the question, but I don't know what I am missing.



The moment generating function of $Y_2$ can be solved using the same method of $Y_1$, but am I right in saying that the moment generating function of K will just be the sum of the moment generating functions of $Y_1$ and $Y_2$?






statistics moment-generating-functions




statistics moment-generating-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 25 at 20:40







peco

















asked Jan 25 at 20:37









pecopeco

888




888












  • $begingroup$
    What is a mgf-pmf?
    $endgroup$
    – Dietrich Burde
    Jan 25 at 20:37










  • $begingroup$
    mgf as in moment generating functions, and pmf as in probability mass function. sorry for the confusion
    $endgroup$
    – peco
    Jan 25 at 20:38










  • $begingroup$
    It should be $e^{-2t}/3+cdots$, not $(e^t/3)^{-2}+cdots$.
    $endgroup$
    – J.G.
    Jan 25 at 23:13


















  • $begingroup$
    What is a mgf-pmf?
    $endgroup$
    – Dietrich Burde
    Jan 25 at 20:37










  • $begingroup$
    mgf as in moment generating functions, and pmf as in probability mass function. sorry for the confusion
    $endgroup$
    – peco
    Jan 25 at 20:38










  • $begingroup$
    It should be $e^{-2t}/3+cdots$, not $(e^t/3)^{-2}+cdots$.
    $endgroup$
    – J.G.
    Jan 25 at 23:13
















$begingroup$
What is a mgf-pmf?
$endgroup$
– Dietrich Burde
Jan 25 at 20:37




$begingroup$
What is a mgf-pmf?
$endgroup$
– Dietrich Burde
Jan 25 at 20:37












$begingroup$
mgf as in moment generating functions, and pmf as in probability mass function. sorry for the confusion
$endgroup$
– peco
Jan 25 at 20:38




$begingroup$
mgf as in moment generating functions, and pmf as in probability mass function. sorry for the confusion
$endgroup$
– peco
Jan 25 at 20:38












$begingroup$
It should be $e^{-2t}/3+cdots$, not $(e^t/3)^{-2}+cdots$.
$endgroup$
– J.G.
Jan 25 at 23:13




$begingroup$
It should be $e^{-2t}/3+cdots$, not $(e^t/3)^{-2}+cdots$.
$endgroup$
– J.G.
Jan 25 at 23:13










1 Answer
1






active

oldest

votes


















0












$begingroup$

If I understand correctly, your $Y_1$ is discrete uniform over the three values ${-2,-1,0}$ and your $Y_2$ is discrete uniform over two values ${1,6}$.



The moment generating function for $Y_1$ is indeed
$$M_1(t) = sum_{y_1 = -2}^0 e^{t y_1} p_1(y_1) = frac13 e^{-2t} + frac13 e^{-t} + frac13 e^{0} = frac13left( 1 + e^{-t} + e^{-2t}right)$$
You just made the mistake of carrying the power over to the probability mass.



The moment generating function for $Y_2$ is similarly
$$M_2(t) = sum_{y_2 = 1,6} e^{t y_2} p_2(y_2) = frac12 e^{-t} + frac12 e^{-6t}= frac12 e^{-t} left(1 + e^{-5t}right)$$



Due to independence, the moment generating function of the sum $K equiv Y_1 + Y_2$ is the product of the respective MGFs.
begin{align}
M_K(t) &= mathbb{E}bigl[ e^{ t(Y_1 + Y_2) }bigr] \
&= mathbb{E}bigl[ e^{ t Y_1 } bigr] cdot mathbb{E}bigl[ e^{ tY_2 } bigr] qquad because Y_1 perp Y_2 \
&= M_1(t) cdot M_2(t) \
&= frac16 e^{-t} left( 1 + e^{-t} + e^{-2t}right)left(1 + e^{-5t}right)end{align}
The terms ca be expanded or rearranged however one prefers.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks, I wouldn't have realized the mistake!
    $endgroup$
    – peco
    Jan 25 at 23:45











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1 Answer
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1 Answer
1






active

oldest

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active

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active

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votes









0












$begingroup$

If I understand correctly, your $Y_1$ is discrete uniform over the three values ${-2,-1,0}$ and your $Y_2$ is discrete uniform over two values ${1,6}$.



The moment generating function for $Y_1$ is indeed
$$M_1(t) = sum_{y_1 = -2}^0 e^{t y_1} p_1(y_1) = frac13 e^{-2t} + frac13 e^{-t} + frac13 e^{0} = frac13left( 1 + e^{-t} + e^{-2t}right)$$
You just made the mistake of carrying the power over to the probability mass.



The moment generating function for $Y_2$ is similarly
$$M_2(t) = sum_{y_2 = 1,6} e^{t y_2} p_2(y_2) = frac12 e^{-t} + frac12 e^{-6t}= frac12 e^{-t} left(1 + e^{-5t}right)$$



Due to independence, the moment generating function of the sum $K equiv Y_1 + Y_2$ is the product of the respective MGFs.
begin{align}
M_K(t) &= mathbb{E}bigl[ e^{ t(Y_1 + Y_2) }bigr] \
&= mathbb{E}bigl[ e^{ t Y_1 } bigr] cdot mathbb{E}bigl[ e^{ tY_2 } bigr] qquad because Y_1 perp Y_2 \
&= M_1(t) cdot M_2(t) \
&= frac16 e^{-t} left( 1 + e^{-t} + e^{-2t}right)left(1 + e^{-5t}right)end{align}
The terms ca be expanded or rearranged however one prefers.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks, I wouldn't have realized the mistake!
    $endgroup$
    – peco
    Jan 25 at 23:45
















0












$begingroup$

If I understand correctly, your $Y_1$ is discrete uniform over the three values ${-2,-1,0}$ and your $Y_2$ is discrete uniform over two values ${1,6}$.



The moment generating function for $Y_1$ is indeed
$$M_1(t) = sum_{y_1 = -2}^0 e^{t y_1} p_1(y_1) = frac13 e^{-2t} + frac13 e^{-t} + frac13 e^{0} = frac13left( 1 + e^{-t} + e^{-2t}right)$$
You just made the mistake of carrying the power over to the probability mass.



The moment generating function for $Y_2$ is similarly
$$M_2(t) = sum_{y_2 = 1,6} e^{t y_2} p_2(y_2) = frac12 e^{-t} + frac12 e^{-6t}= frac12 e^{-t} left(1 + e^{-5t}right)$$



Due to independence, the moment generating function of the sum $K equiv Y_1 + Y_2$ is the product of the respective MGFs.
begin{align}
M_K(t) &= mathbb{E}bigl[ e^{ t(Y_1 + Y_2) }bigr] \
&= mathbb{E}bigl[ e^{ t Y_1 } bigr] cdot mathbb{E}bigl[ e^{ tY_2 } bigr] qquad because Y_1 perp Y_2 \
&= M_1(t) cdot M_2(t) \
&= frac16 e^{-t} left( 1 + e^{-t} + e^{-2t}right)left(1 + e^{-5t}right)end{align}
The terms ca be expanded or rearranged however one prefers.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks, I wouldn't have realized the mistake!
    $endgroup$
    – peco
    Jan 25 at 23:45














0












0








0





$begingroup$

If I understand correctly, your $Y_1$ is discrete uniform over the three values ${-2,-1,0}$ and your $Y_2$ is discrete uniform over two values ${1,6}$.



The moment generating function for $Y_1$ is indeed
$$M_1(t) = sum_{y_1 = -2}^0 e^{t y_1} p_1(y_1) = frac13 e^{-2t} + frac13 e^{-t} + frac13 e^{0} = frac13left( 1 + e^{-t} + e^{-2t}right)$$
You just made the mistake of carrying the power over to the probability mass.



The moment generating function for $Y_2$ is similarly
$$M_2(t) = sum_{y_2 = 1,6} e^{t y_2} p_2(y_2) = frac12 e^{-t} + frac12 e^{-6t}= frac12 e^{-t} left(1 + e^{-5t}right)$$



Due to independence, the moment generating function of the sum $K equiv Y_1 + Y_2$ is the product of the respective MGFs.
begin{align}
M_K(t) &= mathbb{E}bigl[ e^{ t(Y_1 + Y_2) }bigr] \
&= mathbb{E}bigl[ e^{ t Y_1 } bigr] cdot mathbb{E}bigl[ e^{ tY_2 } bigr] qquad because Y_1 perp Y_2 \
&= M_1(t) cdot M_2(t) \
&= frac16 e^{-t} left( 1 + e^{-t} + e^{-2t}right)left(1 + e^{-5t}right)end{align}
The terms ca be expanded or rearranged however one prefers.






share|cite|improve this answer











$endgroup$



If I understand correctly, your $Y_1$ is discrete uniform over the three values ${-2,-1,0}$ and your $Y_2$ is discrete uniform over two values ${1,6}$.



The moment generating function for $Y_1$ is indeed
$$M_1(t) = sum_{y_1 = -2}^0 e^{t y_1} p_1(y_1) = frac13 e^{-2t} + frac13 e^{-t} + frac13 e^{0} = frac13left( 1 + e^{-t} + e^{-2t}right)$$
You just made the mistake of carrying the power over to the probability mass.



The moment generating function for $Y_2$ is similarly
$$M_2(t) = sum_{y_2 = 1,6} e^{t y_2} p_2(y_2) = frac12 e^{-t} + frac12 e^{-6t}= frac12 e^{-t} left(1 + e^{-5t}right)$$



Due to independence, the moment generating function of the sum $K equiv Y_1 + Y_2$ is the product of the respective MGFs.
begin{align}
M_K(t) &= mathbb{E}bigl[ e^{ t(Y_1 + Y_2) }bigr] \
&= mathbb{E}bigl[ e^{ t Y_1 } bigr] cdot mathbb{E}bigl[ e^{ tY_2 } bigr] qquad because Y_1 perp Y_2 \
&= M_1(t) cdot M_2(t) \
&= frac16 e^{-t} left( 1 + e^{-t} + e^{-2t}right)left(1 + e^{-5t}right)end{align}
The terms ca be expanded or rearranged however one prefers.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 25 at 23:25

























answered Jan 25 at 23:01









Lee David Chung LinLee David Chung Lin

4,39531242




4,39531242












  • $begingroup$
    Thanks, I wouldn't have realized the mistake!
    $endgroup$
    – peco
    Jan 25 at 23:45


















  • $begingroup$
    Thanks, I wouldn't have realized the mistake!
    $endgroup$
    – peco
    Jan 25 at 23:45
















$begingroup$
Thanks, I wouldn't have realized the mistake!
$endgroup$
– peco
Jan 25 at 23:45




$begingroup$
Thanks, I wouldn't have realized the mistake!
$endgroup$
– peco
Jan 25 at 23:45


















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