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Is there any way to test a moment generating function_ [closed]

1

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Like isnt there way to put the function value inside and then I would get the same value as the uniform probability function?

probability moment-generating-functions

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asked Mar 25 at 16:45

Jane DoeJane Doe

411

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closed as off-topic by Lee David Chung Lin, zz20s, Eevee Trainer, Thomas Shelby, Leucippus Mar 26 at 5:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lee David Chung Lin, zz20s, Eevee Trainer, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

1

$begingroup$

Like isnt there way to put the function value inside and then I would get the same value as the uniform probability function?

probability moment-generating-functions

share|cite|improve this question

asked Mar 25 at 16:45

Jane DoeJane Doe

411

$endgroup$

closed as off-topic by Lee David Chung Lin, zz20s, Eevee Trainer, Thomas Shelby, Leucippus Mar 26 at 5:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lee David Chung Lin, zz20s, Eevee Trainer, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

$begingroup$

Like isnt there way to put the function value inside and then I would get the same value as the uniform probability function?

probability moment-generating-functions

share|cite|improve this question

asked Mar 25 at 16:45

Jane DoeJane Doe

411

$endgroup$

share|cite|improve this question

asked Mar 25 at 16:45

Jane DoeJane Doe

411

share|cite|improve this question

asked Mar 25 at 16:45

Jane DoeJane Doe

411

asked Mar 25 at 16:45

Jane DoeJane Doe

411

asked Mar 25 at 16:45

Jane DoeJane Doe

411

1 Answer
1

active

oldest

votes

1

$begingroup$

The mgf is given by
$$
M_X(t) = mathbb{E}left[e^{tX}right] = int_mathbb{R} e^{tx} f(x) dx,
$$
so $M_X(0) = 1$ and $$M_X'(0) = mathbb{E}[X]$$ which is a quick shortcut you can use to check things. In your case, $mathbb{E}[X] = 1/2$ by symmetry.

But if you want a formal calculation, your way is correct, namely that
$$
M_X(t) = int_mathbb{R} e^{tx} f(x) dx = int_0^1e^{tx} dx = frac{e^t-1}{t}
$$
The above spot-check is easy to make since
$$
M_X'(t) = e^t t^{-1} - left(e^t-1right)t^{-2} = frac{(t-1)e^t+1}{t^2}
$$
and it's easy to see by L'Hospital's Rule (twice) that $M_X'(t) to 1/2$ as $t to 0$...

share|cite|improve this answer

answered Mar 25 at 17:13

gt6989bgt6989b

36k22557

$endgroup$

add a comment | 

1

$begingroup$

The mgf is given by
$$
M_X(t) = mathbb{E}left[e^{tX}right] = int_mathbb{R} e^{tx} f(x) dx,
$$
so $M_X(0) = 1$ and $$M_X'(0) = mathbb{E}[X]$$ which is a quick shortcut you can use to check things. In your case, $mathbb{E}[X] = 1/2$ by symmetry.

But if you want a formal calculation, your way is correct, namely that
$$
M_X(t) = int_mathbb{R} e^{tx} f(x) dx = int_0^1e^{tx} dx = frac{e^t-1}{t}
$$
The above spot-check is easy to make since
$$
M_X'(t) = e^t t^{-1} - left(e^t-1right)t^{-2} = frac{(t-1)e^t+1}{t^2}
$$
and it's easy to see by L'Hospital's Rule (twice) that $M_X'(t) to 1/2$ as $t to 0$...

share|cite|improve this answer

answered Mar 25 at 17:13

gt6989bgt6989b

36k22557

$endgroup$

add a comment | 

1

$begingroup$

The mgf is given by
$$
M_X(t) = mathbb{E}left[e^{tX}right] = int_mathbb{R} e^{tx} f(x) dx,
$$
so $M_X(0) = 1$ and $$M_X'(0) = mathbb{E}[X]$$ which is a quick shortcut you can use to check things. In your case, $mathbb{E}[X] = 1/2$ by symmetry.

But if you want a formal calculation, your way is correct, namely that
$$
M_X(t) = int_mathbb{R} e^{tx} f(x) dx = int_0^1e^{tx} dx = frac{e^t-1}{t}
$$
The above spot-check is easy to make since
$$
M_X'(t) = e^t t^{-1} - left(e^t-1right)t^{-2} = frac{(t-1)e^t+1}{t^2}
$$
and it's easy to see by L'Hospital's Rule (twice) that $M_X'(t) to 1/2$ as $t to 0$...

share|cite|improve this answer

answered Mar 25 at 17:13

gt6989bgt6989b

36k22557

$endgroup$

add a comment | 

$begingroup$

The mgf is given by
$$
M_X(t) = mathbb{E}left[e^{tX}right] = int_mathbb{R} e^{tx} f(x) dx,
$$
so $M_X(0) = 1$ and $$M_X'(0) = mathbb{E}[X]$$ which is a quick shortcut you can use to check things. In your case, $mathbb{E}[X] = 1/2$ by symmetry.

But if you want a formal calculation, your way is correct, namely that
$$
M_X(t) = int_mathbb{R} e^{tx} f(x) dx = int_0^1e^{tx} dx = frac{e^t-1}{t}
$$
The above spot-check is easy to make since
$$
M_X'(t) = e^t t^{-1} - left(e^t-1right)t^{-2} = frac{(t-1)e^t+1}{t^2}
$$
and it's easy to see by L'Hospital's Rule (twice) that $M_X'(t) to 1/2$ as $t to 0$...

share|cite|improve this answer

answered Mar 25 at 17:13

gt6989bgt6989b

36k22557

$endgroup$

share|cite|improve this answer

answered Mar 25 at 17:13

gt6989bgt6989b

36k22557

answered Mar 25 at 17:13

gt6989bgt6989b

36k22557

answered Mar 25 at 17:13

gt6989bgt6989b

36k22557