Mixing
Transformation of the limits in an integral (From Probability)
$begingroup$
Hello, I am solving the problem above and am close to the solution but I have a small wall that I cannot pass through.
My work is the following.
$$E left[e^{t left(frac{x-mu}{sigma} right)} right] = int_{-infty}^{infty} e^{t left(frac{x-mu}{sigma} right)} f(x) dx$$
$$= e^{ frac{-mu t}{sigma} } int_{-infty}^{infty} e^{frac{tx}{sigma} } f(x) dx$$
$$= e^{ frac{-mu t}{sigma} } M left[frac{t}{sigma}right]$$
But I am not comfortable with why the limit of $t$ becomes $-hsigma < t < hsigma$.
I know that if $$int_{-h}^{h} e^{tx} f(x) dx = M[t]$$
then
$$int_{-h sigma}^{h sigma} e^{frac{tx}{sigma} } f(x) dx = Mleft[frac{t}{sigma}right]$$
from transformation of the variables but I am not confident...
May I have some help, please?
calculus probability statistics
asked Jan 29 at 23:21
hyg17hyg17
2,00222044
$endgroup$
add a comment |
$begingroup$
Hello, I am solving the problem above and am close to the solution but I have a small wall that I cannot pass through.
My work is the following.
$$E left[e^{t left(frac{x-mu}{sigma} right)} right] = int_{-infty}^{infty} e^{t left(frac{x-mu}{sigma} right)} f(x) dx$$
$$= e^{ frac{-mu t}{sigma} } int_{-infty}^{infty} e^{frac{tx}{sigma} } f(x) dx$$
$$= e^{ frac{-mu t}{sigma} } M left[frac{t}{sigma}right]$$
But I am not comfortable with why the limit of $t$ becomes $-hsigma < t < hsigma$.
I know that if $$int_{-h}^{h} e^{tx} f(x) dx = M[t]$$
then
$$int_{-h sigma}^{h sigma} e^{frac{tx}{sigma} } f(x) dx = Mleft[frac{t}{sigma}right]$$
from transformation of the variables but I am not confident...
May I have some help, please?
calculus probability statistics
asked Jan 29 at 23:21
hyg17hyg17
2,00222044
$endgroup$
add a comment |
$begingroup$
Hello, I am solving the problem above and am close to the solution but I have a small wall that I cannot pass through.
My work is the following.
$$E left[e^{t left(frac{x-mu}{sigma} right)} right] = int_{-infty}^{infty} e^{t left(frac{x-mu}{sigma} right)} f(x) dx$$
$$= e^{ frac{-mu t}{sigma} } int_{-infty}^{infty} e^{frac{tx}{sigma} } f(x) dx$$
$$= e^{ frac{-mu t}{sigma} } M left[frac{t}{sigma}right]$$
But I am not comfortable with why the limit of $t$ becomes $-hsigma < t < hsigma$.
I know that if $$int_{-h}^{h} e^{tx} f(x) dx = M[t]$$
then
$$int_{-h sigma}^{h sigma} e^{frac{tx}{sigma} } f(x) dx = Mleft[frac{t}{sigma}right]$$
from transformation of the variables but I am not confident...
May I have some help, please?
calculus probability statistics
asked Jan 29 at 23:21
hyg17hyg17
2,00222044
$endgroup$
Hello, I am solving the problem above and am close to the solution but I have a small wall that I cannot pass through.
My work is the following.
$$E left[e^{t left(frac{x-mu}{sigma} right)} right] = int_{-infty}^{infty} e^{t left(frac{x-mu}{sigma} right)} f(x) dx$$
$$= e^{ frac{-mu t}{sigma} } int_{-infty}^{infty} e^{frac{tx}{sigma} } f(x) dx$$
$$= e^{ frac{-mu t}{sigma} } M left[frac{t}{sigma}right]$$
But I am not comfortable with why the limit of $t$ becomes $-hsigma < t < hsigma$.
I know that if $$int_{-h}^{h} e^{tx} f(x) dx = M[t]$$
then
$$int_{-h sigma}^{h sigma} e^{frac{tx}{sigma} } f(x) dx = Mleft[frac{t}{sigma}right]$$
from transformation of the variables but I am not confident...
May I have some help, please?
calculus probability statistics
calculus probability statistics
asked Jan 29 at 23:21
hyg17hyg17
2,00222044
asked Jan 29 at 23:21
hyg17hyg17
2,00222044
asked Jan 29 at 23:21
hyg17hyg17
2,00222044
asked Jan 29 at 23:21
hyg17hyg17
2,00222044
asked Jan 29 at 23:21
hyg17hyg17
2,00222044
2,00222044
add a comment |
add a comment |
1 Answer
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You can (and should) avoid writing integrals altogether because you don't know if the random variable $X$ has a density.
Imitating your steps yields
$$E e^{t(X-mu)/sigma)} = e^{-t mu / sigma} E[e^{(t/sigma)X}] = e^{-t mu / sigma} M(t/sigma)$$
as long as $M(t/sigma)$ is defined. Since $M(cdot)$ is only defined on the interval $(-h, h)$, we must have $-h < t/sigma < h$.
answered Jan 29 at 23:34
angryavianangryavian
42.5k23481
$endgroup$
$begingroup$
I do not know why I did not notice that... Thank you very much!
$endgroup$
– hyg17
Jan 29 at 23:39
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
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active
oldest
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active
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votes
$begingroup$
You can (and should) avoid writing integrals altogether because you don't know if the random variable $X$ has a density.
Imitating your steps yields
$$E e^{t(X-mu)/sigma)} = e^{-t mu / sigma} E[e^{(t/sigma)X}] = e^{-t mu / sigma} M(t/sigma)$$
as long as $M(t/sigma)$ is defined. Since $M(cdot)$ is only defined on the interval $(-h, h)$, we must have $-h < t/sigma < h$.
answered Jan 29 at 23:34
angryavianangryavian
42.5k23481
$endgroup$
$begingroup$
I do not know why I did not notice that... Thank you very much!
$endgroup$
– hyg17
Jan 29 at 23:39
add a comment |
$begingroup$
You can (and should) avoid writing integrals altogether because you don't know if the random variable $X$ has a density.
Imitating your steps yields
$$E e^{t(X-mu)/sigma)} = e^{-t mu / sigma} E[e^{(t/sigma)X}] = e^{-t mu / sigma} M(t/sigma)$$
as long as $M(t/sigma)$ is defined. Since $M(cdot)$ is only defined on the interval $(-h, h)$, we must have $-h < t/sigma < h$.
answered Jan 29 at 23:34
angryavianangryavian
42.5k23481
$endgroup$
$begingroup$
I do not know why I did not notice that... Thank you very much!
$endgroup$
– hyg17
Jan 29 at 23:39
add a comment |
$begingroup$
You can (and should) avoid writing integrals altogether because you don't know if the random variable $X$ has a density.
Imitating your steps yields
$$E e^{t(X-mu)/sigma)} = e^{-t mu / sigma} E[e^{(t/sigma)X}] = e^{-t mu / sigma} M(t/sigma)$$
as long as $M(t/sigma)$ is defined. Since $M(cdot)$ is only defined on the interval $(-h, h)$, we must have $-h < t/sigma < h$.
answered Jan 29 at 23:34
angryavianangryavian
42.5k23481
$endgroup$
You can (and should) avoid writing integrals altogether because you don't know if the random variable $X$ has a density.
Imitating your steps yields
$$E e^{t(X-mu)/sigma)} = e^{-t mu / sigma} E[e^{(t/sigma)X}] = e^{-t mu / sigma} M(t/sigma)$$
as long as $M(t/sigma)$ is defined. Since $M(cdot)$ is only defined on the interval $(-h, h)$, we must have $-h < t/sigma < h$.
answered Jan 29 at 23:34
angryavianangryavian
42.5k23481
answered Jan 29 at 23:34
angryavianangryavian
42.5k23481
answered Jan 29 at 23:34
angryavianangryavian
42.5k23481
answered Jan 29 at 23:34
angryavianangryavian
42.5k23481
42.5k23481
$begingroup$
I do not know why I did not notice that... Thank you very much!
$endgroup$
– hyg17
Jan 29 at 23:39
add a comment |
$begingroup$
I do not know why I did not notice that... Thank you very much!
$endgroup$
– hyg17
Jan 29 at 23:39
$begingroup$
I do not know why I did not notice that... Thank you very much!
$endgroup$
– hyg17
Jan 29 at 23:39
$begingroup$
I do not know why I did not notice that... Thank you very much!
$endgroup$
– hyg17
Jan 29 at 23:39
add a comment |
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