Finding the Law of X
up vote
0
down vote
favorite
I am woking on a problem and am unsure how to approach it.
It is finding the law of X (the distribution that correspond to) for MGF $2e^tover3-e^t$
I am thinking that this is perhaps an exponential distribution
X ~ Exp(lambda) or perhaps X ~ Poi(lambda)
Do we need to break this apart into:
2e$^t$ and $1over3-e^t$
probability moment-generating-functions moment-problem
asked Nov 18 at 0:00
Ethan
9212
add a comment |
up vote
0
down vote
favorite
I am woking on a problem and am unsure how to approach it.
It is finding the law of X (the distribution that correspond to) for MGF $2e^tover3-e^t$
I am thinking that this is perhaps an exponential distribution
X ~ Exp(lambda) or perhaps X ~ Poi(lambda)
Do we need to break this apart into:
2e$^t$ and $1over3-e^t$
probability moment-generating-functions moment-problem
asked Nov 18 at 0:00
Ethan
9212
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am woking on a problem and am unsure how to approach it.
It is finding the law of X (the distribution that correspond to) for MGF $2e^tover3-e^t$
I am thinking that this is perhaps an exponential distribution
X ~ Exp(lambda) or perhaps X ~ Poi(lambda)
Do we need to break this apart into:
2e$^t$ and $1over3-e^t$
probability moment-generating-functions moment-problem
asked Nov 18 at 0:00
Ethan
9212
I am woking on a problem and am unsure how to approach it.
It is finding the law of X (the distribution that correspond to) for MGF $2e^tover3-e^t$
I am thinking that this is perhaps an exponential distribution
X ~ Exp(lambda) or perhaps X ~ Poi(lambda)
Do we need to break this apart into:
2e$^t$ and $1over3-e^t$
probability moment-generating-functions moment-problem
probability moment-generating-functions moment-problem
asked Nov 18 at 0:00
Ethan
9212
asked Nov 18 at 0:00
Ethan
9212
edited Nov 18 at 0:06
asked Nov 18 at 0:00
Ethan
9212
asked Nov 18 at 0:00
Ethan
9212
asked Nov 18 at 0:00
Ethan
9212
9212
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Use the formula for a geometric series namely $(1-x)^{-1}=sum_{k=0}^infty x^k$ if $|x|<1$, to write that
$$
Ee^{tX}=frac{2e^t}{3-e^t}=2frac{e^t/3}{1-(e^t/3)}=sum_{k=1}^infty2left(frac{e^t}{3}right)^k=sum_{k=1}^infty2times 3^{-k}e^{tk}
$$
provided that $e^t<3$. In any case it follows that
$$
P(X=k)=2times 3^{-k}quad (k=1,2,dotsc)
$$
answered Nov 18 at 0:10
Foobaz John
19.9k41250
Is this the distribution?
– Ethan
Nov 19 at 21:21
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Use the formula for a geometric series namely $(1-x)^{-1}=sum_{k=0}^infty x^k$ if $|x|<1$, to write that
$$
Ee^{tX}=frac{2e^t}{3-e^t}=2frac{e^t/3}{1-(e^t/3)}=sum_{k=1}^infty2left(frac{e^t}{3}right)^k=sum_{k=1}^infty2times 3^{-k}e^{tk}
$$
provided that $e^t<3$. In any case it follows that
$$
P(X=k)=2times 3^{-k}quad (k=1,2,dotsc)
$$
answered Nov 18 at 0:10
Foobaz John
19.9k41250
Is this the distribution?
– Ethan
Nov 19 at 21:21
add a comment |
up vote
0
down vote
Use the formula for a geometric series namely $(1-x)^{-1}=sum_{k=0}^infty x^k$ if $|x|<1$, to write that
$$
Ee^{tX}=frac{2e^t}{3-e^t}=2frac{e^t/3}{1-(e^t/3)}=sum_{k=1}^infty2left(frac{e^t}{3}right)^k=sum_{k=1}^infty2times 3^{-k}e^{tk}
$$
provided that $e^t<3$. In any case it follows that
$$
P(X=k)=2times 3^{-k}quad (k=1,2,dotsc)
$$
answered Nov 18 at 0:10
Foobaz John
19.9k41250
Is this the distribution?
– Ethan
Nov 19 at 21:21
add a comment |
up vote
0
down vote
up vote
0
down vote
Use the formula for a geometric series namely $(1-x)^{-1}=sum_{k=0}^infty x^k$ if $|x|<1$, to write that
$$
Ee^{tX}=frac{2e^t}{3-e^t}=2frac{e^t/3}{1-(e^t/3)}=sum_{k=1}^infty2left(frac{e^t}{3}right)^k=sum_{k=1}^infty2times 3^{-k}e^{tk}
$$
provided that $e^t<3$. In any case it follows that
$$
P(X=k)=2times 3^{-k}quad (k=1,2,dotsc)
$$
answered Nov 18 at 0:10
Foobaz John
19.9k41250
Use the formula for a geometric series namely $(1-x)^{-1}=sum_{k=0}^infty x^k$ if $|x|<1$, to write that
$$
Ee^{tX}=frac{2e^t}{3-e^t}=2frac{e^t/3}{1-(e^t/3)}=sum_{k=1}^infty2left(frac{e^t}{3}right)^k=sum_{k=1}^infty2times 3^{-k}e^{tk}
$$
provided that $e^t<3$. In any case it follows that
$$
P(X=k)=2times 3^{-k}quad (k=1,2,dotsc)
$$
answered Nov 18 at 0:10
Foobaz John
19.9k41250
answered Nov 18 at 0:10
Foobaz John
19.9k41250
answered Nov 18 at 0:10
Foobaz John
19.9k41250
answered Nov 18 at 0:10
Foobaz John
19.9k41250
19.9k41250
Is this the distribution?
– Ethan
Nov 19 at 21:21
add a comment |
Is this the distribution?
– Ethan
Nov 19 at 21:21
Is this the distribution?
– Ethan
Nov 19 at 21:21
Is this the distribution?
– Ethan
Nov 19 at 21:21
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3002984%2ffinding-the-law-of-x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
