Mixing
Moment Generating Function - Cauchy Random Variable
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I want to show that, for every $tneq 0$, the moment generating function of the standard Cauchy distribution is equal to $+infty$, i.e.
$$M_X(t) = int_{-infty}^{+infty} frac{e^{tx}}{1+x^2}dx = +infty$$
I'm really rusty in term of probabilities and integrals, but here is my first try, could anyone validate (or not) this?
begin{align*}
int_{-infty}^{+infty} frac{e^{tx}}{1+x^2}dx &= int_{-infty}^{0} frac{e^{tx}}{1+x^2}dx + int_{0}^{+infty} frac{e^{tx}}{1+x^2}dx \
& geq int_{0}^{+infty} frac{(e^{x})^t}{1+x^2}dx\
& geq int_{0}^{+infty} frac{1}{1+x^2}dx\
&= lim_{xrightarrow +infty} arctan(x)\
&= +infty
end{align*}
probability proof-verification moment-generating-functions
asked Jan 24 at 15:15
Thomas LesgourguesThomas Lesgourgues
1,128119
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add a comment |
$begingroup$
I want to show that, for every $tneq 0$, the moment generating function of the standard Cauchy distribution is equal to $+infty$, i.e.
$$M_X(t) = int_{-infty}^{+infty} frac{e^{tx}}{1+x^2}dx = +infty$$
I'm really rusty in term of probabilities and integrals, but here is my first try, could anyone validate (or not) this?
begin{align*}
int_{-infty}^{+infty} frac{e^{tx}}{1+x^2}dx &= int_{-infty}^{0} frac{e^{tx}}{1+x^2}dx + int_{0}^{+infty} frac{e^{tx}}{1+x^2}dx \
& geq int_{0}^{+infty} frac{(e^{x})^t}{1+x^2}dx\
& geq int_{0}^{+infty} frac{1}{1+x^2}dx\
&= lim_{xrightarrow +infty} arctan(x)\
&= +infty
end{align*}
probability proof-verification moment-generating-functions
asked Jan 24 at 15:15
Thomas LesgourguesThomas Lesgourgues
1,128119
$endgroup$
$begingroup$
That last equality doesn't look right.
$endgroup$
– StubbornAtom
Jan 24 at 18:02
$begingroup$
Indeed!! I went way too fast on that one.
$endgroup$
– Thomas Lesgourgues
Jan 24 at 18:12
$begingroup$
The fact that the Cauchy distribution does not have finite moments of order $ge 1$ is an adequate argument for the nonexistence of the moment generating function.
$endgroup$
– StubbornAtom
Jan 25 at 18:08
add a comment |
$begingroup$
I want to show that, for every $tneq 0$, the moment generating function of the standard Cauchy distribution is equal to $+infty$, i.e.
$$M_X(t) = int_{-infty}^{+infty} frac{e^{tx}}{1+x^2}dx = +infty$$
I'm really rusty in term of probabilities and integrals, but here is my first try, could anyone validate (or not) this?
begin{align*}
int_{-infty}^{+infty} frac{e^{tx}}{1+x^2}dx &= int_{-infty}^{0} frac{e^{tx}}{1+x^2}dx + int_{0}^{+infty} frac{e^{tx}}{1+x^2}dx \
& geq int_{0}^{+infty} frac{(e^{x})^t}{1+x^2}dx\
& geq int_{0}^{+infty} frac{1}{1+x^2}dx\
&= lim_{xrightarrow +infty} arctan(x)\
&= +infty
end{align*}
probability proof-verification moment-generating-functions
asked Jan 24 at 15:15
Thomas LesgourguesThomas Lesgourgues
1,128119
$endgroup$
I want to show that, for every $tneq 0$, the moment generating function of the standard Cauchy distribution is equal to $+infty$, i.e.
$$M_X(t) = int_{-infty}^{+infty} frac{e^{tx}}{1+x^2}dx = +infty$$
I'm really rusty in term of probabilities and integrals, but here is my first try, could anyone validate (or not) this?
begin{align*}
int_{-infty}^{+infty} frac{e^{tx}}{1+x^2}dx &= int_{-infty}^{0} frac{e^{tx}}{1+x^2}dx + int_{0}^{+infty} frac{e^{tx}}{1+x^2}dx \
& geq int_{0}^{+infty} frac{(e^{x})^t}{1+x^2}dx\
& geq int_{0}^{+infty} frac{1}{1+x^2}dx\
&= lim_{xrightarrow +infty} arctan(x)\
&= +infty
end{align*}
probability proof-verification moment-generating-functions
probability proof-verification moment-generating-functions
asked Jan 24 at 15:15
Thomas LesgourguesThomas Lesgourgues
1,128119
asked Jan 24 at 15:15
Thomas LesgourguesThomas Lesgourgues
1,128119
edited Jan 24 at 15:32
Thomas Lesgourgues
asked Jan 24 at 15:15
Thomas LesgourguesThomas Lesgourgues
1,128119
asked Jan 24 at 15:15
Thomas LesgourguesThomas Lesgourgues
1,128119
asked Jan 24 at 15:15
Thomas LesgourguesThomas Lesgourgues
1,128119
1,128119
$begingroup$
That last equality doesn't look right.
$endgroup$
– StubbornAtom
Jan 24 at 18:02
$begingroup$
Indeed!! I went way too fast on that one.
$endgroup$
– Thomas Lesgourgues
Jan 24 at 18:12
$begingroup$
The fact that the Cauchy distribution does not have finite moments of order $ge 1$ is an adequate argument for the nonexistence of the moment generating function.
$endgroup$
– StubbornAtom
Jan 25 at 18:08
add a comment |
$begingroup$
That last equality doesn't look right.
$endgroup$
– StubbornAtom
Jan 24 at 18:02
$begingroup$
Indeed!! I went way too fast on that one.
$endgroup$
– Thomas Lesgourgues
Jan 24 at 18:12
$begingroup$
The fact that the Cauchy distribution does not have finite moments of order $ge 1$ is an adequate argument for the nonexistence of the moment generating function.
$endgroup$
– StubbornAtom
Jan 25 at 18:08
$begingroup$
That last equality doesn't look right.
$endgroup$
– StubbornAtom
Jan 24 at 18:02
$begingroup$
That last equality doesn't look right.
$endgroup$
– StubbornAtom
Jan 24 at 18:02
$begingroup$
Indeed!! I went way too fast on that one.
$endgroup$
– Thomas Lesgourgues
Jan 24 at 18:12
$begingroup$
Indeed!! I went way too fast on that one.
$endgroup$
– Thomas Lesgourgues
Jan 24 at 18:12
$begingroup$
The fact that the Cauchy distribution does not have finite moments of order $ge 1$ is an adequate argument for the nonexistence of the moment generating function.
$endgroup$
– StubbornAtom
Jan 25 at 18:08
$begingroup$
The fact that the Cauchy distribution does not have finite moments of order $ge 1$ is an adequate argument for the nonexistence of the moment generating function.
$endgroup$
– StubbornAtom
Jan 25 at 18:08
add a comment |
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$begingroup$
That last equality doesn't look right.
$endgroup$
– StubbornAtom
Jan 24 at 18:02
$begingroup$
Indeed!! I went way too fast on that one.
$endgroup$
– Thomas Lesgourgues
Jan 24 at 18:12
$begingroup$
The fact that the Cauchy distribution does not have finite moments of order $ge 1$ is an adequate argument for the nonexistence of the moment generating function.
$endgroup$
– StubbornAtom
Jan 25 at 18:08