Mixing
Find the Distribution that corresponds to the given MGF
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I am working on a problem and am a little bit confused.
I need to find the distribution that corresponds to the MGF:
$2e^tover3-e^t$
Do we need to separate this into something like:
2e$^t$ and $1over3-e^t$
It seems like we might be able to separate this into X ~ Exp() or Poi()
probability moment-generating-functions moment-problem
asked 2 days ago
Ethan
9212
add a comment |
up vote
0
down vote
favorite
I am working on a problem and am a little bit confused.
I need to find the distribution that corresponds to the MGF:
$2e^tover3-e^t$
Do we need to separate this into something like:
2e$^t$ and $1over3-e^t$
It seems like we might be able to separate this into X ~ Exp() or Poi()
probability moment-generating-functions moment-problem
asked 2 days ago
Ethan
9212
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am working on a problem and am a little bit confused.
I need to find the distribution that corresponds to the MGF:
$2e^tover3-e^t$
Do we need to separate this into something like:
2e$^t$ and $1over3-e^t$
It seems like we might be able to separate this into X ~ Exp() or Poi()
probability moment-generating-functions moment-problem
asked 2 days ago
Ethan
9212
I am working on a problem and am a little bit confused.
I need to find the distribution that corresponds to the MGF:
$2e^tover3-e^t$
Do we need to separate this into something like:
2e$^t$ and $1over3-e^t$
It seems like we might be able to separate this into X ~ Exp() or Poi()
probability moment-generating-functions moment-problem
probability moment-generating-functions moment-problem
asked 2 days ago
Ethan
9212
asked 2 days ago
Ethan
9212
asked 2 days ago
Ethan
9212
asked 2 days ago
Ethan
9212
asked 2 days ago
Ethan
9212
9212
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The MGF of a Poisson r.v. features a double exponential, and that of an exponential r.v. is a rational function (no exponential), so it looks unlikely at first glance these two would be candidates (also, unclear what you mean by "separate" here.)
However, eyeballing the Wikipedia table showing a list of standard MGFs, you can spot one that looks quite similar, that of the geometric distribution. Namely, if $X$ follows a Geometric distribution with parameter $p$, then the MGF of X is
$$
M_x(t) = frac{p e^t}{1-(1-p)e^t}, quad t < logfrac{1}{1-p} tag{1}
$$
So let's see. What you have is
$$
M_X(t) = frac{2e^t}{3-e^t} = frac{frac{2}{3}e^t}{1-frac{1}{3}e^t}
= frac{frac{2}{3}e^t}{1-left(1-frac{2}{3}right)e^t} tag{2}
$$
which should allow you to conclude.
answered 2 days ago
Clement C.
48.7k33784
Thank you this clears a lot of things up.
– Ethan
2 days ago
@Ethan You're welcome!
– Clement C.
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The MGF of a Poisson r.v. features a double exponential, and that of an exponential r.v. is a rational function (no exponential), so it looks unlikely at first glance these two would be candidates (also, unclear what you mean by "separate" here.)
However, eyeballing the Wikipedia table showing a list of standard MGFs, you can spot one that looks quite similar, that of the geometric distribution. Namely, if $X$ follows a Geometric distribution with parameter $p$, then the MGF of X is
$$
M_x(t) = frac{p e^t}{1-(1-p)e^t}, quad t < logfrac{1}{1-p} tag{1}
$$
So let's see. What you have is
$$
M_X(t) = frac{2e^t}{3-e^t} = frac{frac{2}{3}e^t}{1-frac{1}{3}e^t}
= frac{frac{2}{3}e^t}{1-left(1-frac{2}{3}right)e^t} tag{2}
$$
which should allow you to conclude.
answered 2 days ago
Clement C.
48.7k33784
Thank you this clears a lot of things up.
– Ethan
2 days ago
@Ethan You're welcome!
– Clement C.
2 days ago
add a comment |
up vote
1
down vote
accepted
The MGF of a Poisson r.v. features a double exponential, and that of an exponential r.v. is a rational function (no exponential), so it looks unlikely at first glance these two would be candidates (also, unclear what you mean by "separate" here.)
However, eyeballing the Wikipedia table showing a list of standard MGFs, you can spot one that looks quite similar, that of the geometric distribution. Namely, if $X$ follows a Geometric distribution with parameter $p$, then the MGF of X is
$$
M_x(t) = frac{p e^t}{1-(1-p)e^t}, quad t < logfrac{1}{1-p} tag{1}
$$
So let's see. What you have is
$$
M_X(t) = frac{2e^t}{3-e^t} = frac{frac{2}{3}e^t}{1-frac{1}{3}e^t}
= frac{frac{2}{3}e^t}{1-left(1-frac{2}{3}right)e^t} tag{2}
$$
which should allow you to conclude.
answered 2 days ago
Clement C.
48.7k33784
Thank you this clears a lot of things up.
– Ethan
2 days ago
@Ethan You're welcome!
– Clement C.
2 days ago
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The MGF of a Poisson r.v. features a double exponential, and that of an exponential r.v. is a rational function (no exponential), so it looks unlikely at first glance these two would be candidates (also, unclear what you mean by "separate" here.)
However, eyeballing the Wikipedia table showing a list of standard MGFs, you can spot one that looks quite similar, that of the geometric distribution. Namely, if $X$ follows a Geometric distribution with parameter $p$, then the MGF of X is
$$
M_x(t) = frac{p e^t}{1-(1-p)e^t}, quad t < logfrac{1}{1-p} tag{1}
$$
So let's see. What you have is
$$
M_X(t) = frac{2e^t}{3-e^t} = frac{frac{2}{3}e^t}{1-frac{1}{3}e^t}
= frac{frac{2}{3}e^t}{1-left(1-frac{2}{3}right)e^t} tag{2}
$$
which should allow you to conclude.
answered 2 days ago
Clement C.
48.7k33784
The MGF of a Poisson r.v. features a double exponential, and that of an exponential r.v. is a rational function (no exponential), so it looks unlikely at first glance these two would be candidates (also, unclear what you mean by "separate" here.)
However, eyeballing the Wikipedia table showing a list of standard MGFs, you can spot one that looks quite similar, that of the geometric distribution. Namely, if $X$ follows a Geometric distribution with parameter $p$, then the MGF of X is
$$
M_x(t) = frac{p e^t}{1-(1-p)e^t}, quad t < logfrac{1}{1-p} tag{1}
$$
So let's see. What you have is
$$
M_X(t) = frac{2e^t}{3-e^t} = frac{frac{2}{3}e^t}{1-frac{1}{3}e^t}
= frac{frac{2}{3}e^t}{1-left(1-frac{2}{3}right)e^t} tag{2}
$$
which should allow you to conclude.
answered 2 days ago
Clement C.
48.7k33784
answered 2 days ago
Clement C.
48.7k33784
answered 2 days ago
Clement C.
48.7k33784
answered 2 days ago
Clement C.
48.7k33784
48.7k33784
Thank you this clears a lot of things up.
– Ethan
2 days ago
@Ethan You're welcome!
– Clement C.
2 days ago
add a comment |
Thank you this clears a lot of things up.
– Ethan
2 days ago
@Ethan You're welcome!
– Clement C.
2 days ago
Thank you this clears a lot of things up.
– Ethan
2 days ago
Thank you this clears a lot of things up.
– Ethan
2 days ago
@Ethan You're welcome!
– Clement C.
2 days ago
@Ethan You're welcome!
– Clement C.
2 days ago
add a comment |
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