Mixing
Characteristic function : in what is more useful than the density function?
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The principal thing I know with the characteristic function is : $X$ and $Y$ have the same law $iff$ the characteristic function are the same. But if they have the same density function, they also have the same law, no ? So $X$ and $Y$ has the same law $iff$ they have the same density function. Now, in what the characteristic function is more useful than the density to prove that two r.v. have the same law ? Since we first need to know that the density function before computing the characteristic function.
probability
edited Jan 18 at 21:03
David C. Ullrich
61k43994
asked Jan 18 at 18:08
NewMathNewMath
4059
$endgroup$
add a comment |
$begingroup$
The principal thing I know with the characteristic function is : $X$ and $Y$ have the same law $iff$ the characteristic function are the same. But if they have the same density function, they also have the same law, no ? So $X$ and $Y$ has the same law $iff$ they have the same density function. Now, in what the characteristic function is more useful than the density to prove that two r.v. have the same law ? Since we first need to know that the density function before computing the characteristic function.
probability
edited Jan 18 at 21:03
David C. Ullrich
61k43994
asked Jan 18 at 18:08
NewMathNewMath
4059
$endgroup$
2
$begingroup$
The characteristic function is used for much more than that. Wait tiil you see the proof of the Central Limit Theorem...
$endgroup$
– David C. Ullrich
Jan 18 at 21:01
1
$begingroup$
Not all random variables have density functions! And there can be many ways to determine the characteristic function of a random variable, other than by doing an integral with respect to its density function (assuming it even has one).
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– Nate Eldredge
Jan 18 at 21:13
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@NateEldredge: How can you determine the characteristic function to a r.v. that has no density function since the characteristic function is the Fourier transform of the density function. Could you provide an example ?
$endgroup$
– NewMath
Jan 18 at 21:32
1
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@NewMath: The characteristic function is defined as $phi(t) = E[e^{itX}]$. So for instance, if $X$ has a Bernoulli(1/2) distribution (which is discrete and has no density), we have $phi(t) = frac{1}{2} e^{it cdot 0} + frac{1}{2} e^{it cdot 1} = frac{1}{2} + frac{1}{2} e^{it}$.
$endgroup$
– Nate Eldredge
Jan 18 at 21:37
add a comment |
$begingroup$
The principal thing I know with the characteristic function is : $X$ and $Y$ have the same law $iff$ the characteristic function are the same. But if they have the same density function, they also have the same law, no ? So $X$ and $Y$ has the same law $iff$ they have the same density function. Now, in what the characteristic function is more useful than the density to prove that two r.v. have the same law ? Since we first need to know that the density function before computing the characteristic function.
probability
edited Jan 18 at 21:03
David C. Ullrich
61k43994
asked Jan 18 at 18:08
NewMathNewMath
4059
$endgroup$
The principal thing I know with the characteristic function is : $X$ and $Y$ have the same law $iff$ the characteristic function are the same. But if they have the same density function, they also have the same law, no ? So $X$ and $Y$ has the same law $iff$ they have the same density function. Now, in what the characteristic function is more useful than the density to prove that two r.v. have the same law ? Since we first need to know that the density function before computing the characteristic function.
probability
probability
edited Jan 18 at 21:03
David C. Ullrich
61k43994
asked Jan 18 at 18:08
NewMathNewMath
4059
edited Jan 18 at 21:03
David C. Ullrich
61k43994
asked Jan 18 at 18:08
NewMathNewMath
4059
edited Jan 18 at 21:03
David C. Ullrich
61k43994
edited Jan 18 at 21:03
David C. Ullrich
61k43994
edited Jan 18 at 21:03
David C. Ullrich
61k43994
61k43994
asked Jan 18 at 18:08
NewMathNewMath
4059
asked Jan 18 at 18:08
NewMathNewMath
4059
asked Jan 18 at 18:08
NewMathNewMath
4059
4059
2
$begingroup$
The characteristic function is used for much more than that. Wait tiil you see the proof of the Central Limit Theorem...
$endgroup$
– David C. Ullrich
Jan 18 at 21:01
1
$begingroup$
Not all random variables have density functions! And there can be many ways to determine the characteristic function of a random variable, other than by doing an integral with respect to its density function (assuming it even has one).
$endgroup$
– Nate Eldredge
Jan 18 at 21:13
$begingroup$
@NateEldredge: How can you determine the characteristic function to a r.v. that has no density function since the characteristic function is the Fourier transform of the density function. Could you provide an example ?
$endgroup$
– NewMath
Jan 18 at 21:32
1
$begingroup$
@NewMath: The characteristic function is defined as $phi(t) = E[e^{itX}]$. So for instance, if $X$ has a Bernoulli(1/2) distribution (which is discrete and has no density), we have $phi(t) = frac{1}{2} e^{it cdot 0} + frac{1}{2} e^{it cdot 1} = frac{1}{2} + frac{1}{2} e^{it}$.
$endgroup$
– Nate Eldredge
Jan 18 at 21:37
add a comment |
2
$begingroup$
The characteristic function is used for much more than that. Wait tiil you see the proof of the Central Limit Theorem...
$endgroup$
– David C. Ullrich
Jan 18 at 21:01
1
$begingroup$
Not all random variables have density functions! And there can be many ways to determine the characteristic function of a random variable, other than by doing an integral with respect to its density function (assuming it even has one).
$endgroup$
– Nate Eldredge
Jan 18 at 21:13
$begingroup$
@NateEldredge: How can you determine the characteristic function to a r.v. that has no density function since the characteristic function is the Fourier transform of the density function. Could you provide an example ?
$endgroup$
– NewMath
Jan 18 at 21:32
1
$begingroup$
@NewMath: The characteristic function is defined as $phi(t) = E[e^{itX}]$. So for instance, if $X$ has a Bernoulli(1/2) distribution (which is discrete and has no density), we have $phi(t) = frac{1}{2} e^{it cdot 0} + frac{1}{2} e^{it cdot 1} = frac{1}{2} + frac{1}{2} e^{it}$.
$endgroup$
– Nate Eldredge
Jan 18 at 21:37
2
2
$begingroup$
The characteristic function is used for much more than that. Wait tiil you see the proof of the Central Limit Theorem...
$endgroup$
– David C. Ullrich
Jan 18 at 21:01
$begingroup$
The characteristic function is used for much more than that. Wait tiil you see the proof of the Central Limit Theorem...
$endgroup$
– David C. Ullrich
Jan 18 at 21:01
1
1
$begingroup$
Not all random variables have density functions! And there can be many ways to determine the characteristic function of a random variable, other than by doing an integral with respect to its density function (assuming it even has one).
$endgroup$
– Nate Eldredge
Jan 18 at 21:13
$begingroup$
Not all random variables have density functions! And there can be many ways to determine the characteristic function of a random variable, other than by doing an integral with respect to its density function (assuming it even has one).
$endgroup$
– Nate Eldredge
Jan 18 at 21:13
$begingroup$
@NateEldredge: How can you determine the characteristic function to a r.v. that has no density function since the characteristic function is the Fourier transform of the density function. Could you provide an example ?
$endgroup$
– NewMath
Jan 18 at 21:32
$begingroup$
@NateEldredge: How can you determine the characteristic function to a r.v. that has no density function since the characteristic function is the Fourier transform of the density function. Could you provide an example ?
$endgroup$
– NewMath
Jan 18 at 21:32
1
1
$begingroup$
@NewMath: The characteristic function is defined as $phi(t) = E[e^{itX}]$. So for instance, if $X$ has a Bernoulli(1/2) distribution (which is discrete and has no density), we have $phi(t) = frac{1}{2} e^{it cdot 0} + frac{1}{2} e^{it cdot 1} = frac{1}{2} + frac{1}{2} e^{it}$.
$endgroup$
– Nate Eldredge
Jan 18 at 21:37
$begingroup$
@NewMath: The characteristic function is defined as $phi(t) = E[e^{itX}]$. So for instance, if $X$ has a Bernoulli(1/2) distribution (which is discrete and has no density), we have $phi(t) = frac{1}{2} e^{it cdot 0} + frac{1}{2} e^{it cdot 1} = frac{1}{2} + frac{1}{2} e^{it}$.
$endgroup$
– Nate Eldredge
Jan 18 at 21:37
add a comment |
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$begingroup$
The characteristic function is used for much more than that. Wait tiil you see the proof of the Central Limit Theorem...
$endgroup$
– David C. Ullrich
Jan 18 at 21:01
1
$begingroup$
Not all random variables have density functions! And there can be many ways to determine the characteristic function of a random variable, other than by doing an integral with respect to its density function (assuming it even has one).
$endgroup$
– Nate Eldredge
Jan 18 at 21:13
$begingroup$
@NateEldredge: How can you determine the characteristic function to a r.v. that has no density function since the characteristic function is the Fourier transform of the density function. Could you provide an example ?
$endgroup$
– NewMath
Jan 18 at 21:32
1
$begingroup$
@NewMath: The characteristic function is defined as $phi(t) = E[e^{itX}]$. So for instance, if $X$ has a Bernoulli(1/2) distribution (which is discrete and has no density), we have $phi(t) = frac{1}{2} e^{it cdot 0} + frac{1}{2} e^{it cdot 1} = frac{1}{2} + frac{1}{2} e^{it}$.
$endgroup$
– Nate Eldredge
Jan 18 at 21:37