Is $gamma(t)=cos^2(t)$ a characteristic function?
-1
$begingroup$
Is $gamma(t)=cos^2(t)$ a characteristic function?
I don't know how can I show that $gamma(t)=cos^2(t)$ is a positive function.
probability probability-theory characteristic-functions
edited Nov 4 '14 at 22:31
mookid
25.6k52447
asked Nov 4 '14 at 21:57
Adams_20Adams_20
1
$endgroup$
add a comment |
-1
$begingroup$
Is $gamma(t)=cos^2(t)$ a characteristic function?
I don't know how can I show that $gamma(t)=cos^2(t)$ is a positive function.
probability probability-theory characteristic-functions
edited Nov 4 '14 at 22:31
mookid
25.6k52447
asked Nov 4 '14 at 21:57
Adams_20Adams_20
1
$endgroup$
1
$begingroup$
Do you know what the characteristic function of a finite distribution looks like? Like, say, a coin flip (i.e. even odds Bernoulli distribution)?
$endgroup$
– Arthur
Nov 4 '14 at 22:04
$begingroup$
I know that, in this function we can compute: $X ~~~ (2,1/4), (-2,1/4), (0,1/2)$
$endgroup$
– Adams_20
Nov 4 '14 at 22:07
1
$begingroup$
$cos^2 t = dfrac{1+cos 2t}{2}=dfrac{1}{2}e^{0}+...$ and $cos x= dfrac{e^{ix}+e^{-ix}}{2}$
$endgroup$
– Cortizol
Nov 4 '14 at 22:09
add a comment |
-1
-1
-1
$begingroup$
Is $gamma(t)=cos^2(t)$ a characteristic function?
I don't know how can I show that $gamma(t)=cos^2(t)$ is a positive function.
probability probability-theory characteristic-functions
edited Nov 4 '14 at 22:31
mookid
25.6k52447
asked Nov 4 '14 at 21:57
Adams_20Adams_20
1
$endgroup$
Is $gamma(t)=cos^2(t)$ a characteristic function?
I don't know how can I show that $gamma(t)=cos^2(t)$ is a positive function.
probability probability-theory characteristic-functions
probability probability-theory characteristic-functions
edited Nov 4 '14 at 22:31
mookid
25.6k52447
asked Nov 4 '14 at 21:57
Adams_20Adams_20
1
edited Nov 4 '14 at 22:31
mookid
25.6k52447
asked Nov 4 '14 at 21:57
Adams_20Adams_20
1
edited Nov 4 '14 at 22:31
mookid
25.6k52447
edited Nov 4 '14 at 22:31
mookid
25.6k52447
edited Nov 4 '14 at 22:31
mookid
25.6k52447
25.6k52447
asked Nov 4 '14 at 21:57
Adams_20Adams_20
1
asked Nov 4 '14 at 21:57
Adams_20Adams_20
1
asked Nov 4 '14 at 21:57
Adams_20Adams_20
1
1
1
$begingroup$
Do you know what the characteristic function of a finite distribution looks like? Like, say, a coin flip (i.e. even odds Bernoulli distribution)?
$endgroup$
– Arthur
Nov 4 '14 at 22:04
$begingroup$
I know that, in this function we can compute: $X ~~~ (2,1/4), (-2,1/4), (0,1/2)$
$endgroup$
– Adams_20
Nov 4 '14 at 22:07
1
$begingroup$
$cos^2 t = dfrac{1+cos 2t}{2}=dfrac{1}{2}e^{0}+...$ and $cos x= dfrac{e^{ix}+e^{-ix}}{2}$
$endgroup$
– Cortizol
Nov 4 '14 at 22:09
add a comment |
1
$begingroup$
Do you know what the characteristic function of a finite distribution looks like? Like, say, a coin flip (i.e. even odds Bernoulli distribution)?
$endgroup$
– Arthur
Nov 4 '14 at 22:04
$begingroup$
I know that, in this function we can compute: $X ~~~ (2,1/4), (-2,1/4), (0,1/2)$
$endgroup$
– Adams_20
Nov 4 '14 at 22:07
1
$begingroup$
$cos^2 t = dfrac{1+cos 2t}{2}=dfrac{1}{2}e^{0}+...$ and $cos x= dfrac{e^{ix}+e^{-ix}}{2}$
$endgroup$
– Cortizol
Nov 4 '14 at 22:09
1
1
$begingroup$
Do you know what the characteristic function of a finite distribution looks like? Like, say, a coin flip (i.e. even odds Bernoulli distribution)?
$endgroup$
– Arthur
Nov 4 '14 at 22:04
$begingroup$
Do you know what the characteristic function of a finite distribution looks like? Like, say, a coin flip (i.e. even odds Bernoulli distribution)?
$endgroup$
– Arthur
Nov 4 '14 at 22:04
$begingroup$
I know that, in this function we can compute: $X ~~~ (2,1/4), (-2,1/4), (0,1/2)$
$endgroup$
– Adams_20
Nov 4 '14 at 22:07
$begingroup$
I know that, in this function we can compute: $X ~~~ (2,1/4), (-2,1/4), (0,1/2)$
$endgroup$
– Adams_20
Nov 4 '14 at 22:07
1
1
$begingroup$
$cos^2 t = dfrac{1+cos 2t}{2}=dfrac{1}{2}e^{0}+...$ and $cos x= dfrac{e^{ix}+e^{-ix}}{2}$
$endgroup$
– Cortizol
Nov 4 '14 at 22:09
$begingroup$
$cos^2 t = dfrac{1+cos 2t}{2}=dfrac{1}{2}e^{0}+...$ and $cos x= dfrac{e^{ix}+e^{-ix}}{2}$
$endgroup$
– Cortizol
Nov 4 '14 at 22:09
add a comment |
1 Answer
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Use and show the following facts:
- if $X$ and $Y$ are two independent random variables, the characteristic function of $X+Y$ is the product of the characteristic functions of $X$ with that of $Y$;
- the function $tmapsto cos t$ is a characteristic function.
answered Nov 5 '14 at 11:38
Davide GiraudoDavide Giraudo
126k16150261
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add a comment |
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0
$begingroup$
Use and show the following facts:
- if $X$ and $Y$ are two independent random variables, the characteristic function of $X+Y$ is the product of the characteristic functions of $X$ with that of $Y$;
- the function $tmapsto cos t$ is a characteristic function.
answered Nov 5 '14 at 11:38
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
add a comment |
0
$begingroup$
Use and show the following facts:
- if $X$ and $Y$ are two independent random variables, the characteristic function of $X+Y$ is the product of the characteristic functions of $X$ with that of $Y$;
- the function $tmapsto cos t$ is a characteristic function.
answered Nov 5 '14 at 11:38
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
add a comment |
0
0
0
$begingroup$
Use and show the following facts:
- if $X$ and $Y$ are two independent random variables, the characteristic function of $X+Y$ is the product of the characteristic functions of $X$ with that of $Y$;
- the function $tmapsto cos t$ is a characteristic function.
answered Nov 5 '14 at 11:38
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
Use and show the following facts:
- if $X$ and $Y$ are two independent random variables, the characteristic function of $X+Y$ is the product of the characteristic functions of $X$ with that of $Y$;
- the function $tmapsto cos t$ is a characteristic function.
answered Nov 5 '14 at 11:38
Davide GiraudoDavide Giraudo
126k16150261
answered Nov 5 '14 at 11:38
Davide GiraudoDavide Giraudo
126k16150261
answered Nov 5 '14 at 11:38
Davide GiraudoDavide Giraudo
126k16150261
answered Nov 5 '14 at 11:38
Davide GiraudoDavide Giraudo
126k16150261
126k16150261
add a comment |
add a comment |
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1
$begingroup$
Do you know what the characteristic function of a finite distribution looks like? Like, say, a coin flip (i.e. even odds Bernoulli distribution)?
$endgroup$
– Arthur
Nov 4 '14 at 22:04
$begingroup$
I know that, in this function we can compute: $X ~~~ (2,1/4), (-2,1/4), (0,1/2)$
$endgroup$
– Adams_20
Nov 4 '14 at 22:07
1
$begingroup$
$cos^2 t = dfrac{1+cos 2t}{2}=dfrac{1}{2}e^{0}+...$ and $cos x= dfrac{e^{ix}+e^{-ix}}{2}$
$endgroup$
– Cortizol
Nov 4 '14 at 22:09