Finding $E(X)$ using moment generating function
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Say:
$m_X(t) = left(frac11-tright)^1/2 cdot left(frac44-tright)^1/2 cdot left(frac99-tright)^1/2 $
We want. $E(X)$.
The only theorem in my textbook is that
$m_X^k(0) = E(X^k)$
As in, the derivative is proportional to the power k to $E(X)$.
Gamma is: $X$ ~ $Gamma(alpha, lambda)$ with mgt $$left(fraclambdalambda-tright)^alpha$$
so $m_X(t) = Gamma(alpha=1/2, lambda = 1) cdot Gamma(alpha=1/2, lambda = 4) cdot Gamma(alpha=1/2, lambda = 9)$.
$E(X) = fracalphalambda$
Solution:
$E(X) = 1/2/1 + 1/2/4 + 1/2/9 = 49/72$
What formula does this use to get above?
probability expectation
asked Aug 10 at 19:58
Bas bas
39611
add a comment |Â
up vote
0
down vote
favorite
Say:
$m_X(t) = left(frac11-tright)^1/2 cdot left(frac44-tright)^1/2 cdot left(frac99-tright)^1/2 $
We want. $E(X)$.
The only theorem in my textbook is that
$m_X^k(0) = E(X^k)$
As in, the derivative is proportional to the power k to $E(X)$.
Gamma is: $X$ ~ $Gamma(alpha, lambda)$ with mgt $$left(fraclambdalambda-tright)^alpha$$
so $m_X(t) = Gamma(alpha=1/2, lambda = 1) cdot Gamma(alpha=1/2, lambda = 4) cdot Gamma(alpha=1/2, lambda = 9)$.
$E(X) = fracalphalambda$
Solution:
$E(X) = 1/2/1 + 1/2/4 + 1/2/9 = 49/72$
What formula does this use to get above?
probability expectation
asked Aug 10 at 19:58
Bas bas
39611
Didn't you already ask exactly this today? To which you received an answer...
– Did
Aug 10 at 20:36
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Say:
$m_X(t) = left(frac11-tright)^1/2 cdot left(frac44-tright)^1/2 cdot left(frac99-tright)^1/2 $
We want. $E(X)$.
The only theorem in my textbook is that
$m_X^k(0) = E(X^k)$
As in, the derivative is proportional to the power k to $E(X)$.
Gamma is: $X$ ~ $Gamma(alpha, lambda)$ with mgt $$left(fraclambdalambda-tright)^alpha$$
so $m_X(t) = Gamma(alpha=1/2, lambda = 1) cdot Gamma(alpha=1/2, lambda = 4) cdot Gamma(alpha=1/2, lambda = 9)$.
$E(X) = fracalphalambda$
Solution:
$E(X) = 1/2/1 + 1/2/4 + 1/2/9 = 49/72$
What formula does this use to get above?
probability expectation
asked Aug 10 at 19:58
Bas bas
39611
Say:
$m_X(t) = left(frac11-tright)^1/2 cdot left(frac44-tright)^1/2 cdot left(frac99-tright)^1/2 $
We want. $E(X)$.
The only theorem in my textbook is that
$m_X^k(0) = E(X^k)$
As in, the derivative is proportional to the power k to $E(X)$.
Gamma is: $X$ ~ $Gamma(alpha, lambda)$ with mgt $$left(fraclambdalambda-tright)^alpha$$
so $m_X(t) = Gamma(alpha=1/2, lambda = 1) cdot Gamma(alpha=1/2, lambda = 4) cdot Gamma(alpha=1/2, lambda = 9)$.
$E(X) = fracalphalambda$
Solution:
$E(X) = 1/2/1 + 1/2/4 + 1/2/9 = 49/72$
What formula does this use to get above?
probability expectation
asked Aug 10 at 19:58
Bas bas
39611
asked Aug 10 at 19:58
Bas bas
39611
asked Aug 10 at 19:58
Bas bas
39611
asked Aug 10 at 19:58
Bas bas
39611
39611
Didn't you already ask exactly this today? To which you received an answer...
– Did
Aug 10 at 20:36
add a comment |Â
Didn't you already ask exactly this today? To which you received an answer...
– Did
Aug 10 at 20:36
Didn't you already ask exactly this today? To which you received an answer...
– Did
Aug 10 at 20:36
Didn't you already ask exactly this today? To which you received an answer...
– Did
Aug 10 at 20:36
add a comment |Â
2 Answers
2
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up vote
1
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The case $k=1$ gives $E(X)=m_X'(0)$. Since $m_X(0)$, $E(X)$ is also the value at $t=0$ of $$dfracddtln m_X=dfrac-12dfracddt(ln (1-t)+ln (4-t)+ln (9-t))=frac12(dfrac11-t+dfrac14-t+dfrac19-t).$$For $t=0$, this gives $tfrac4972$.
answered Aug 10 at 21:08
J.G.
13.5k11424
add a comment |Â
up vote
0
down vote
Hint: Using the theorem, with $k=1$ you get that $m_X'(0) = E(X)$, so find the first derivative of $m_X(t)$ at $0$ to get the answer.
answered Aug 10 at 20:22
B. Mehta
11.7k21944
There is a much simpler route.
– Did
Aug 10 at 20:37
Sure, but the poster appears interested in the approach using the given theorem.
– B. Mehta
Aug 10 at 20:58
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The case $k=1$ gives $E(X)=m_X'(0)$. Since $m_X(0)$, $E(X)$ is also the value at $t=0$ of $$dfracddtln m_X=dfrac-12dfracddt(ln (1-t)+ln (4-t)+ln (9-t))=frac12(dfrac11-t+dfrac14-t+dfrac19-t).$$For $t=0$, this gives $tfrac4972$.
answered Aug 10 at 21:08
J.G.
13.5k11424
add a comment |Â
up vote
1
down vote
The case $k=1$ gives $E(X)=m_X'(0)$. Since $m_X(0)$, $E(X)$ is also the value at $t=0$ of $$dfracddtln m_X=dfrac-12dfracddt(ln (1-t)+ln (4-t)+ln (9-t))=frac12(dfrac11-t+dfrac14-t+dfrac19-t).$$For $t=0$, this gives $tfrac4972$.
answered Aug 10 at 21:08
J.G.
13.5k11424
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The case $k=1$ gives $E(X)=m_X'(0)$. Since $m_X(0)$, $E(X)$ is also the value at $t=0$ of $$dfracddtln m_X=dfrac-12dfracddt(ln (1-t)+ln (4-t)+ln (9-t))=frac12(dfrac11-t+dfrac14-t+dfrac19-t).$$For $t=0$, this gives $tfrac4972$.
answered Aug 10 at 21:08
J.G.
13.5k11424
The case $k=1$ gives $E(X)=m_X'(0)$. Since $m_X(0)$, $E(X)$ is also the value at $t=0$ of $$dfracddtln m_X=dfrac-12dfracddt(ln (1-t)+ln (4-t)+ln (9-t))=frac12(dfrac11-t+dfrac14-t+dfrac19-t).$$For $t=0$, this gives $tfrac4972$.
answered Aug 10 at 21:08
J.G.
13.5k11424
answered Aug 10 at 21:08
J.G.
13.5k11424
answered Aug 10 at 21:08
J.G.
13.5k11424
answered Aug 10 at 21:08
J.G.
13.5k11424
13.5k11424
add a comment |Â
add a comment |Â
up vote
0
down vote
Hint: Using the theorem, with $k=1$ you get that $m_X'(0) = E(X)$, so find the first derivative of $m_X(t)$ at $0$ to get the answer.
answered Aug 10 at 20:22
B. Mehta
11.7k21944
There is a much simpler route.
– Did
Aug 10 at 20:37
Sure, but the poster appears interested in the approach using the given theorem.
– B. Mehta
Aug 10 at 20:58
add a comment |Â
up vote
0
down vote
Hint: Using the theorem, with $k=1$ you get that $m_X'(0) = E(X)$, so find the first derivative of $m_X(t)$ at $0$ to get the answer.
answered Aug 10 at 20:22
B. Mehta
11.7k21944
There is a much simpler route.
– Did
Aug 10 at 20:37
Sure, but the poster appears interested in the approach using the given theorem.
– B. Mehta
Aug 10 at 20:58
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint: Using the theorem, with $k=1$ you get that $m_X'(0) = E(X)$, so find the first derivative of $m_X(t)$ at $0$ to get the answer.
answered Aug 10 at 20:22
B. Mehta
11.7k21944
Hint: Using the theorem, with $k=1$ you get that $m_X'(0) = E(X)$, so find the first derivative of $m_X(t)$ at $0$ to get the answer.
answered Aug 10 at 20:22
B. Mehta
11.7k21944
answered Aug 10 at 20:22
B. Mehta
11.7k21944
answered Aug 10 at 20:22
B. Mehta
11.7k21944
answered Aug 10 at 20:22
B. Mehta
11.7k21944
11.7k21944
There is a much simpler route.
– Did
Aug 10 at 20:37
Sure, but the poster appears interested in the approach using the given theorem.
– B. Mehta
Aug 10 at 20:58
add a comment |Â
There is a much simpler route.
– Did
Aug 10 at 20:37
Sure, but the poster appears interested in the approach using the given theorem.
– B. Mehta
Aug 10 at 20:58
There is a much simpler route.
– Did
Aug 10 at 20:37
There is a much simpler route.
– Did
Aug 10 at 20:37
Sure, but the poster appears interested in the approach using the given theorem.
– B. Mehta
Aug 10 at 20:58
Sure, but the poster appears interested in the approach using the given theorem.
– B. Mehta
Aug 10 at 20:58
add a comment |Â
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Didn't you already ask exactly this today? To which you received an answer...
– Did
Aug 10 at 20:36