Mixing
Moment Generating function of one variable and Probability
$begingroup$
I need only a hint please :
Is there a way to calculate the probability if we have the moment generating function as in the following question:
IF the moment generating function is given by :
$$M(t)= e^{4.6(e^t-1)}$$
Find $P(3<X<6)$
What I have tried is :
Suppose $S = {s_1,s_2,dots}$ then
$$e^{4.6(e^t-1)}=M(t)=sum_{s_iin S}e^{tx}f(s_i)= e^{ts_1}f(s_1)+e^{ts_2}f(s_2)+cdots $$
$$e^{4.6e^t}=e^{4.6}left( e^{ts_1}f(s_1)+e^{ts_2}f(s_2)+cdots right)$$
If we write
$$e^{4.6e^t}=1+(4.6e^t)+frac{(4.6e^t)^2}{2}+cdots$$
Then
$$1+(4.6e^t)+frac{(4.6e^t)^2}{2}+cdots = e^{(ts_1+4.6)}f(s_1)+e^{(ts_2+4.6)}f(s_2)+cdots$$
So
$$1+4.6e^t+10.58e^{2t}+cdots = e^{(ts_1+4.6)}f(s_1)+e^{(ts_2+4.6)}f(s_2)+cdots$$
probability moment-generating-functions
asked Jan 27 at 20:34
AhmedAhmed
1,247612
$endgroup$
add a comment |
$begingroup$
I need only a hint please :
Is there a way to calculate the probability if we have the moment generating function as in the following question:
IF the moment generating function is given by :
$$M(t)= e^{4.6(e^t-1)}$$
Find $P(3<X<6)$
What I have tried is :
Suppose $S = {s_1,s_2,dots}$ then
$$e^{4.6(e^t-1)}=M(t)=sum_{s_iin S}e^{tx}f(s_i)= e^{ts_1}f(s_1)+e^{ts_2}f(s_2)+cdots $$
$$e^{4.6e^t}=e^{4.6}left( e^{ts_1}f(s_1)+e^{ts_2}f(s_2)+cdots right)$$
If we write
$$e^{4.6e^t}=1+(4.6e^t)+frac{(4.6e^t)^2}{2}+cdots$$
Then
$$1+(4.6e^t)+frac{(4.6e^t)^2}{2}+cdots = e^{(ts_1+4.6)}f(s_1)+e^{(ts_2+4.6)}f(s_2)+cdots$$
So
$$1+4.6e^t+10.58e^{2t}+cdots = e^{(ts_1+4.6)}f(s_1)+e^{(ts_2+4.6)}f(s_2)+cdots$$
probability moment-generating-functions
asked Jan 27 at 20:34
AhmedAhmed
1,247612
$endgroup$
$begingroup$
I edited the question @kimchilover
$endgroup$
– Ahmed
Jan 28 at 4:09
$begingroup$
I have edited the post with what I have tried @kimchilover
$endgroup$
– Ahmed
Jan 28 at 4:35
add a comment |
$begingroup$
I need only a hint please :
Is there a way to calculate the probability if we have the moment generating function as in the following question:
IF the moment generating function is given by :
$$M(t)= e^{4.6(e^t-1)}$$
Find $P(3<X<6)$
What I have tried is :
Suppose $S = {s_1,s_2,dots}$ then
$$e^{4.6(e^t-1)}=M(t)=sum_{s_iin S}e^{tx}f(s_i)= e^{ts_1}f(s_1)+e^{ts_2}f(s_2)+cdots $$
$$e^{4.6e^t}=e^{4.6}left( e^{ts_1}f(s_1)+e^{ts_2}f(s_2)+cdots right)$$
If we write
$$e^{4.6e^t}=1+(4.6e^t)+frac{(4.6e^t)^2}{2}+cdots$$
Then
$$1+(4.6e^t)+frac{(4.6e^t)^2}{2}+cdots = e^{(ts_1+4.6)}f(s_1)+e^{(ts_2+4.6)}f(s_2)+cdots$$
So
$$1+4.6e^t+10.58e^{2t}+cdots = e^{(ts_1+4.6)}f(s_1)+e^{(ts_2+4.6)}f(s_2)+cdots$$
probability moment-generating-functions
asked Jan 27 at 20:34
AhmedAhmed
1,247612
$endgroup$
I need only a hint please :
Is there a way to calculate the probability if we have the moment generating function as in the following question:
IF the moment generating function is given by :
$$M(t)= e^{4.6(e^t-1)}$$
Find $P(3<X<6)$
What I have tried is :
Suppose $S = {s_1,s_2,dots}$ then
$$e^{4.6(e^t-1)}=M(t)=sum_{s_iin S}e^{tx}f(s_i)= e^{ts_1}f(s_1)+e^{ts_2}f(s_2)+cdots $$
$$e^{4.6e^t}=e^{4.6}left( e^{ts_1}f(s_1)+e^{ts_2}f(s_2)+cdots right)$$
If we write
$$e^{4.6e^t}=1+(4.6e^t)+frac{(4.6e^t)^2}{2}+cdots$$
Then
$$1+(4.6e^t)+frac{(4.6e^t)^2}{2}+cdots = e^{(ts_1+4.6)}f(s_1)+e^{(ts_2+4.6)}f(s_2)+cdots$$
So
$$1+4.6e^t+10.58e^{2t}+cdots = e^{(ts_1+4.6)}f(s_1)+e^{(ts_2+4.6)}f(s_2)+cdots$$
probability moment-generating-functions
probability moment-generating-functions
asked Jan 27 at 20:34
AhmedAhmed
1,247612
asked Jan 27 at 20:34
AhmedAhmed
1,247612
edited Jan 28 at 4:34
Ahmed
asked Jan 27 at 20:34
AhmedAhmed
1,247612
asked Jan 27 at 20:34
AhmedAhmed
1,247612
asked Jan 27 at 20:34
AhmedAhmed
1,247612
1,247612
$begingroup$
I edited the question @kimchilover
$endgroup$
– Ahmed
Jan 28 at 4:09
$begingroup$
I have edited the post with what I have tried @kimchilover
$endgroup$
– Ahmed
Jan 28 at 4:35
add a comment |
$begingroup$
I edited the question @kimchilover
$endgroup$
– Ahmed
Jan 28 at 4:09
$begingroup$
I have edited the post with what I have tried @kimchilover
$endgroup$
– Ahmed
Jan 28 at 4:35
$begingroup$
I edited the question @kimchilover
$endgroup$
– Ahmed
Jan 28 at 4:09
$begingroup$
I edited the question @kimchilover
$endgroup$
– Ahmed
Jan 28 at 4:09
$begingroup$
I have edited the post with what I have tried @kimchilover
$endgroup$
– Ahmed
Jan 28 at 4:35
$begingroup$
I have edited the post with what I have tried @kimchilover
$endgroup$
– Ahmed
Jan 28 at 4:35
add a comment |
1 Answer
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active
oldest
votes
$begingroup$
The requested probability $P(3 < X < 6)$ has nothing to do with how the moment-generating function (henceforth MGF) is defined or any of its properties. These kinds of exercises simply intend to ask you to recognize a common distribution from the given MGF (look-up tables are available in most textbooks).
Having said that, in this particular case, your approach happens to work as a practical solution.
Denote the given MGF as $M(t) = e^{lambda (e^t - 1)}$ with $lambda = 4.6$ so that your last equation becomes
$$ 1 + lambda e^t + frac{ lambda^2 e^{2t} }{ 2! } + frac{ lambda^3 e^{3t} }{ 3! } + cdots = e^{(ts_1 + lambda)}f(s_1) + e^{(ts_2 + lambda )}f(s_2) + e^{(ts_3 + lambda )}f(s_3) + cdots $$
From which one can make a first round of guess as the following.
Since the terms $f(s_k)$ of probability mass function (henceforth PMF) don't carry any $t$:
Naive Guess: $ s_1 = 0,~ s_2 = 1,~ s_3 = 2,~ ldots$ etc to match the $e^0,e^t,e^{2t},ldots$ etc. Namely, each $s_k = k-1$ to match each $e^{(k-1)t}$.
With this guess, the support is $k = 0,1,2,3,ldots$ the non-negative integers. The desired probability thus becomes $P(3 < X < 6) = P(4 leq X leq 5)$.
Note that $s_1 = 0$ together with $1 = e^{(ts_1 + lambda)}f(s_1)$ gives $f(s_1) = e^{-lambda}$ at $k = 1$. As for other $s_k~$, we have the ratio between adjacent terms
begin{align}
&&frac{ e^{(t s_k + lambda )}f(s_k) }{ e^{(ts_{k-1} + lambda )} f(s_{k-1}) } & = frac{ lambda^{k-1} e^{(k -1)t} / (k-1)!}{ lambda^{k-2} e^{(k-2)t} / (k-2)! } \
implies&& frac{ f(s_k) }{ f(s_{k-1}) } &= frac{lambda e^{t(s_k - s_{k-1})} }{k-1} = frac{lambda e^t }{k-1} quad because s_k - s_{k-1} = 1 ~~ forall,kgeq 1
end{align}
Therefore, from $f(s_1)$ we have $f(s_2)$, then from $f(s_2)$ comes $f(s_3)$, and so on.
begin{align}
f(s_1) = e^{-lambda} implies && f(s_2) &= f(s_1)frac{lambda e^{t(s_k - s_{k-1})} }{k-1}Bigg|_{k = 2} = e^{-lambda} \
implies && f(s_3) &= e^{-lambda} frac{ e^{2t} }2 \
implies && f(s_4) &= e^{-lambda} frac{ e^{3t} }{ 3!} \
vdots&&&
end{align}
This is the Poisson distribution, and our work is done because the MGF is unique. The first round of naive guess that $s_k = k-1$ turns out to be right on, and there's no need for a second round of improvement or anything.
The requested probability $displaystyle P(4 leq X leq 5) = e^{-lambda} frac{lambda^4}{4!} bigl( 1 + frac{ lambda }5bigr)$ with $lambda = 4.6$.
answered Jan 28 at 10:57
Lee David Chung LinLee David Chung Lin
4,45341242
$endgroup$
add a comment |
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$begingroup$
The requested probability $P(3 < X < 6)$ has nothing to do with how the moment-generating function (henceforth MGF) is defined or any of its properties. These kinds of exercises simply intend to ask you to recognize a common distribution from the given MGF (look-up tables are available in most textbooks).
Having said that, in this particular case, your approach happens to work as a practical solution.
Denote the given MGF as $M(t) = e^{lambda (e^t - 1)}$ with $lambda = 4.6$ so that your last equation becomes
$$ 1 + lambda e^t + frac{ lambda^2 e^{2t} }{ 2! } + frac{ lambda^3 e^{3t} }{ 3! } + cdots = e^{(ts_1 + lambda)}f(s_1) + e^{(ts_2 + lambda )}f(s_2) + e^{(ts_3 + lambda )}f(s_3) + cdots $$
From which one can make a first round of guess as the following.
Since the terms $f(s_k)$ of probability mass function (henceforth PMF) don't carry any $t$:
Naive Guess: $ s_1 = 0,~ s_2 = 1,~ s_3 = 2,~ ldots$ etc to match the $e^0,e^t,e^{2t},ldots$ etc. Namely, each $s_k = k-1$ to match each $e^{(k-1)t}$.
With this guess, the support is $k = 0,1,2,3,ldots$ the non-negative integers. The desired probability thus becomes $P(3 < X < 6) = P(4 leq X leq 5)$.
Note that $s_1 = 0$ together with $1 = e^{(ts_1 + lambda)}f(s_1)$ gives $f(s_1) = e^{-lambda}$ at $k = 1$. As for other $s_k~$, we have the ratio between adjacent terms
begin{align}
&&frac{ e^{(t s_k + lambda )}f(s_k) }{ e^{(ts_{k-1} + lambda )} f(s_{k-1}) } & = frac{ lambda^{k-1} e^{(k -1)t} / (k-1)!}{ lambda^{k-2} e^{(k-2)t} / (k-2)! } \
implies&& frac{ f(s_k) }{ f(s_{k-1}) } &= frac{lambda e^{t(s_k - s_{k-1})} }{k-1} = frac{lambda e^t }{k-1} quad because s_k - s_{k-1} = 1 ~~ forall,kgeq 1
end{align}
Therefore, from $f(s_1)$ we have $f(s_2)$, then from $f(s_2)$ comes $f(s_3)$, and so on.
begin{align}
f(s_1) = e^{-lambda} implies && f(s_2) &= f(s_1)frac{lambda e^{t(s_k - s_{k-1})} }{k-1}Bigg|_{k = 2} = e^{-lambda} \
implies && f(s_3) &= e^{-lambda} frac{ e^{2t} }2 \
implies && f(s_4) &= e^{-lambda} frac{ e^{3t} }{ 3!} \
vdots&&&
end{align}
This is the Poisson distribution, and our work is done because the MGF is unique. The first round of naive guess that $s_k = k-1$ turns out to be right on, and there's no need for a second round of improvement or anything.
The requested probability $displaystyle P(4 leq X leq 5) = e^{-lambda} frac{lambda^4}{4!} bigl( 1 + frac{ lambda }5bigr)$ with $lambda = 4.6$.
answered Jan 28 at 10:57
Lee David Chung LinLee David Chung Lin
4,45341242
$endgroup$
add a comment |
$begingroup$
The requested probability $P(3 < X < 6)$ has nothing to do with how the moment-generating function (henceforth MGF) is defined or any of its properties. These kinds of exercises simply intend to ask you to recognize a common distribution from the given MGF (look-up tables are available in most textbooks).
Having said that, in this particular case, your approach happens to work as a practical solution.
Denote the given MGF as $M(t) = e^{lambda (e^t - 1)}$ with $lambda = 4.6$ so that your last equation becomes
$$ 1 + lambda e^t + frac{ lambda^2 e^{2t} }{ 2! } + frac{ lambda^3 e^{3t} }{ 3! } + cdots = e^{(ts_1 + lambda)}f(s_1) + e^{(ts_2 + lambda )}f(s_2) + e^{(ts_3 + lambda )}f(s_3) + cdots $$
From which one can make a first round of guess as the following.
Since the terms $f(s_k)$ of probability mass function (henceforth PMF) don't carry any $t$:
Naive Guess: $ s_1 = 0,~ s_2 = 1,~ s_3 = 2,~ ldots$ etc to match the $e^0,e^t,e^{2t},ldots$ etc. Namely, each $s_k = k-1$ to match each $e^{(k-1)t}$.
With this guess, the support is $k = 0,1,2,3,ldots$ the non-negative integers. The desired probability thus becomes $P(3 < X < 6) = P(4 leq X leq 5)$.
Note that $s_1 = 0$ together with $1 = e^{(ts_1 + lambda)}f(s_1)$ gives $f(s_1) = e^{-lambda}$ at $k = 1$. As for other $s_k~$, we have the ratio between adjacent terms
begin{align}
&&frac{ e^{(t s_k + lambda )}f(s_k) }{ e^{(ts_{k-1} + lambda )} f(s_{k-1}) } & = frac{ lambda^{k-1} e^{(k -1)t} / (k-1)!}{ lambda^{k-2} e^{(k-2)t} / (k-2)! } \
implies&& frac{ f(s_k) }{ f(s_{k-1}) } &= frac{lambda e^{t(s_k - s_{k-1})} }{k-1} = frac{lambda e^t }{k-1} quad because s_k - s_{k-1} = 1 ~~ forall,kgeq 1
end{align}
Therefore, from $f(s_1)$ we have $f(s_2)$, then from $f(s_2)$ comes $f(s_3)$, and so on.
begin{align}
f(s_1) = e^{-lambda} implies && f(s_2) &= f(s_1)frac{lambda e^{t(s_k - s_{k-1})} }{k-1}Bigg|_{k = 2} = e^{-lambda} \
implies && f(s_3) &= e^{-lambda} frac{ e^{2t} }2 \
implies && f(s_4) &= e^{-lambda} frac{ e^{3t} }{ 3!} \
vdots&&&
end{align}
This is the Poisson distribution, and our work is done because the MGF is unique. The first round of naive guess that $s_k = k-1$ turns out to be right on, and there's no need for a second round of improvement or anything.
The requested probability $displaystyle P(4 leq X leq 5) = e^{-lambda} frac{lambda^4}{4!} bigl( 1 + frac{ lambda }5bigr)$ with $lambda = 4.6$.
answered Jan 28 at 10:57
Lee David Chung LinLee David Chung Lin
4,45341242
$endgroup$
add a comment |
$begingroup$
The requested probability $P(3 < X < 6)$ has nothing to do with how the moment-generating function (henceforth MGF) is defined or any of its properties. These kinds of exercises simply intend to ask you to recognize a common distribution from the given MGF (look-up tables are available in most textbooks).
Having said that, in this particular case, your approach happens to work as a practical solution.
Denote the given MGF as $M(t) = e^{lambda (e^t - 1)}$ with $lambda = 4.6$ so that your last equation becomes
$$ 1 + lambda e^t + frac{ lambda^2 e^{2t} }{ 2! } + frac{ lambda^3 e^{3t} }{ 3! } + cdots = e^{(ts_1 + lambda)}f(s_1) + e^{(ts_2 + lambda )}f(s_2) + e^{(ts_3 + lambda )}f(s_3) + cdots $$
From which one can make a first round of guess as the following.
Since the terms $f(s_k)$ of probability mass function (henceforth PMF) don't carry any $t$:
Naive Guess: $ s_1 = 0,~ s_2 = 1,~ s_3 = 2,~ ldots$ etc to match the $e^0,e^t,e^{2t},ldots$ etc. Namely, each $s_k = k-1$ to match each $e^{(k-1)t}$.
With this guess, the support is $k = 0,1,2,3,ldots$ the non-negative integers. The desired probability thus becomes $P(3 < X < 6) = P(4 leq X leq 5)$.
Note that $s_1 = 0$ together with $1 = e^{(ts_1 + lambda)}f(s_1)$ gives $f(s_1) = e^{-lambda}$ at $k = 1$. As for other $s_k~$, we have the ratio between adjacent terms
begin{align}
&&frac{ e^{(t s_k + lambda )}f(s_k) }{ e^{(ts_{k-1} + lambda )} f(s_{k-1}) } & = frac{ lambda^{k-1} e^{(k -1)t} / (k-1)!}{ lambda^{k-2} e^{(k-2)t} / (k-2)! } \
implies&& frac{ f(s_k) }{ f(s_{k-1}) } &= frac{lambda e^{t(s_k - s_{k-1})} }{k-1} = frac{lambda e^t }{k-1} quad because s_k - s_{k-1} = 1 ~~ forall,kgeq 1
end{align}
Therefore, from $f(s_1)$ we have $f(s_2)$, then from $f(s_2)$ comes $f(s_3)$, and so on.
begin{align}
f(s_1) = e^{-lambda} implies && f(s_2) &= f(s_1)frac{lambda e^{t(s_k - s_{k-1})} }{k-1}Bigg|_{k = 2} = e^{-lambda} \
implies && f(s_3) &= e^{-lambda} frac{ e^{2t} }2 \
implies && f(s_4) &= e^{-lambda} frac{ e^{3t} }{ 3!} \
vdots&&&
end{align}
This is the Poisson distribution, and our work is done because the MGF is unique. The first round of naive guess that $s_k = k-1$ turns out to be right on, and there's no need for a second round of improvement or anything.
The requested probability $displaystyle P(4 leq X leq 5) = e^{-lambda} frac{lambda^4}{4!} bigl( 1 + frac{ lambda }5bigr)$ with $lambda = 4.6$.
answered Jan 28 at 10:57
Lee David Chung LinLee David Chung Lin
4,45341242
$endgroup$
The requested probability $P(3 < X < 6)$ has nothing to do with how the moment-generating function (henceforth MGF) is defined or any of its properties. These kinds of exercises simply intend to ask you to recognize a common distribution from the given MGF (look-up tables are available in most textbooks).
Having said that, in this particular case, your approach happens to work as a practical solution.
Denote the given MGF as $M(t) = e^{lambda (e^t - 1)}$ with $lambda = 4.6$ so that your last equation becomes
$$ 1 + lambda e^t + frac{ lambda^2 e^{2t} }{ 2! } + frac{ lambda^3 e^{3t} }{ 3! } + cdots = e^{(ts_1 + lambda)}f(s_1) + e^{(ts_2 + lambda )}f(s_2) + e^{(ts_3 + lambda )}f(s_3) + cdots $$
From which one can make a first round of guess as the following.
Since the terms $f(s_k)$ of probability mass function (henceforth PMF) don't carry any $t$:
Naive Guess: $ s_1 = 0,~ s_2 = 1,~ s_3 = 2,~ ldots$ etc to match the $e^0,e^t,e^{2t},ldots$ etc. Namely, each $s_k = k-1$ to match each $e^{(k-1)t}$.
With this guess, the support is $k = 0,1,2,3,ldots$ the non-negative integers. The desired probability thus becomes $P(3 < X < 6) = P(4 leq X leq 5)$.
Note that $s_1 = 0$ together with $1 = e^{(ts_1 + lambda)}f(s_1)$ gives $f(s_1) = e^{-lambda}$ at $k = 1$. As for other $s_k~$, we have the ratio between adjacent terms
begin{align}
&&frac{ e^{(t s_k + lambda )}f(s_k) }{ e^{(ts_{k-1} + lambda )} f(s_{k-1}) } & = frac{ lambda^{k-1} e^{(k -1)t} / (k-1)!}{ lambda^{k-2} e^{(k-2)t} / (k-2)! } \
implies&& frac{ f(s_k) }{ f(s_{k-1}) } &= frac{lambda e^{t(s_k - s_{k-1})} }{k-1} = frac{lambda e^t }{k-1} quad because s_k - s_{k-1} = 1 ~~ forall,kgeq 1
end{align}
Therefore, from $f(s_1)$ we have $f(s_2)$, then from $f(s_2)$ comes $f(s_3)$, and so on.
begin{align}
f(s_1) = e^{-lambda} implies && f(s_2) &= f(s_1)frac{lambda e^{t(s_k - s_{k-1})} }{k-1}Bigg|_{k = 2} = e^{-lambda} \
implies && f(s_3) &= e^{-lambda} frac{ e^{2t} }2 \
implies && f(s_4) &= e^{-lambda} frac{ e^{3t} }{ 3!} \
vdots&&&
end{align}
This is the Poisson distribution, and our work is done because the MGF is unique. The first round of naive guess that $s_k = k-1$ turns out to be right on, and there's no need for a second round of improvement or anything.
The requested probability $displaystyle P(4 leq X leq 5) = e^{-lambda} frac{lambda^4}{4!} bigl( 1 + frac{ lambda }5bigr)$ with $lambda = 4.6$.
answered Jan 28 at 10:57
Lee David Chung LinLee David Chung Lin
4,45341242
edited Jan 28 at 21:57
answered Jan 28 at 10:57
Lee David Chung LinLee David Chung Lin
4,45341242
answered Jan 28 at 10:57
Lee David Chung LinLee David Chung Lin
4,45341242
answered Jan 28 at 10:57
Lee David Chung LinLee David Chung Lin
4,45341242
4,45341242
add a comment |
add a comment |
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$begingroup$
I edited the question @kimchilover
$endgroup$
– Ahmed
Jan 28 at 4:09
$begingroup$
I have edited the post with what I have tried @kimchilover
$endgroup$
– Ahmed
Jan 28 at 4:35