Mixing
Show that X and -X are identically distributed and their moment, $M_{x}(t)$ is even
$begingroup$
Let's assume X is a random variable with an even pdf.
To show that X and -X are identically distributed, we need to prove that $F_X(x)=F_-X(x)$. We also know that X having an even pdf means $f_X(x)=-f_X(x)$. What I did here is that I took the integral of $f_X(x)$ and $-f_X(x)$ both from negative infinity to x and set them equal to each other. When we integrate this we should get $F_X(x)=F_-X(x)$, right?
I'm not so sure how to show that $M_x(t)$, or the moment is even. Do we just do $M_x(t)=-M_x(-t)$ and infer our results from there?
probability statistics moment-generating-functions
asked Sep 27 '15 at 16:11
cambelotcambelot
1,03311334
$endgroup$
add a comment |
$begingroup$
Let's assume X is a random variable with an even pdf.
To show that X and -X are identically distributed, we need to prove that $F_X(x)=F_-X(x)$. We also know that X having an even pdf means $f_X(x)=-f_X(x)$. What I did here is that I took the integral of $f_X(x)$ and $-f_X(x)$ both from negative infinity to x and set them equal to each other. When we integrate this we should get $F_X(x)=F_-X(x)$, right?
I'm not so sure how to show that $M_x(t)$, or the moment is even. Do we just do $M_x(t)=-M_x(-t)$ and infer our results from there?
probability statistics moment-generating-functions
asked Sep 27 '15 at 16:11
cambelotcambelot
1,03311334
$endgroup$
$begingroup$
you mean $f_X(x) = f_X(-x)$
$endgroup$
– Ant
Sep 27 '15 at 17:16
add a comment |
$begingroup$
Let's assume X is a random variable with an even pdf.
To show that X and -X are identically distributed, we need to prove that $F_X(x)=F_-X(x)$. We also know that X having an even pdf means $f_X(x)=-f_X(x)$. What I did here is that I took the integral of $f_X(x)$ and $-f_X(x)$ both from negative infinity to x and set them equal to each other. When we integrate this we should get $F_X(x)=F_-X(x)$, right?
I'm not so sure how to show that $M_x(t)$, or the moment is even. Do we just do $M_x(t)=-M_x(-t)$ and infer our results from there?
probability statistics moment-generating-functions
asked Sep 27 '15 at 16:11
cambelotcambelot
1,03311334
$endgroup$
Let's assume X is a random variable with an even pdf.
To show that X and -X are identically distributed, we need to prove that $F_X(x)=F_-X(x)$. We also know that X having an even pdf means $f_X(x)=-f_X(x)$. What I did here is that I took the integral of $f_X(x)$ and $-f_X(x)$ both from negative infinity to x and set them equal to each other. When we integrate this we should get $F_X(x)=F_-X(x)$, right?
I'm not so sure how to show that $M_x(t)$, or the moment is even. Do we just do $M_x(t)=-M_x(-t)$ and infer our results from there?
probability statistics moment-generating-functions
probability statistics moment-generating-functions
asked Sep 27 '15 at 16:11
cambelotcambelot
1,03311334
asked Sep 27 '15 at 16:11
cambelotcambelot
1,03311334
asked Sep 27 '15 at 16:11
cambelotcambelot
1,03311334
asked Sep 27 '15 at 16:11
cambelotcambelot
1,03311334
asked Sep 27 '15 at 16:11
cambelotcambelot
1,03311334
1,03311334
$begingroup$
you mean $f_X(x) = f_X(-x)$
$endgroup$
– Ant
Sep 27 '15 at 17:16
add a comment |
$begingroup$
you mean $f_X(x) = f_X(-x)$
$endgroup$
– Ant
Sep 27 '15 at 17:16
$begingroup$
you mean $f_X(x) = f_X(-x)$
$endgroup$
– Ant
Sep 27 '15 at 17:16
$begingroup$
you mean $f_X(x) = f_X(-x)$
$endgroup$
– Ant
Sep 27 '15 at 17:16
add a comment |
1 Answer
1
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oldest
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$begingroup$
It looks like you've got the first part OK. For the second part you need to show that $M_X(-t)=M_X(t)$, which you can do by making use of the first part:
begin{eqnarray*}
M_X(-t) &=& ; E(e^{-tx}) \
&=& ; E(e^{t(-x)}) \
&=& ; M_{-X}(t) \
&=& ; M_X(t)qquadqquadtext{since $X$ and $-X$ are identically distributed.}
end{eqnarray*}
answered Sep 27 '15 at 17:13
Mick AMick A
8,8502825
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
It looks like you've got the first part OK. For the second part you need to show that $M_X(-t)=M_X(t)$, which you can do by making use of the first part:
begin{eqnarray*}
M_X(-t) &=& ; E(e^{-tx}) \
&=& ; E(e^{t(-x)}) \
&=& ; M_{-X}(t) \
&=& ; M_X(t)qquadqquadtext{since $X$ and $-X$ are identically distributed.}
end{eqnarray*}
answered Sep 27 '15 at 17:13
Mick AMick A
8,8502825
$endgroup$
add a comment |
$begingroup$
It looks like you've got the first part OK. For the second part you need to show that $M_X(-t)=M_X(t)$, which you can do by making use of the first part:
begin{eqnarray*}
M_X(-t) &=& ; E(e^{-tx}) \
&=& ; E(e^{t(-x)}) \
&=& ; M_{-X}(t) \
&=& ; M_X(t)qquadqquadtext{since $X$ and $-X$ are identically distributed.}
end{eqnarray*}
answered Sep 27 '15 at 17:13
Mick AMick A
8,8502825
$endgroup$
add a comment |
$begingroup$
It looks like you've got the first part OK. For the second part you need to show that $M_X(-t)=M_X(t)$, which you can do by making use of the first part:
begin{eqnarray*}
M_X(-t) &=& ; E(e^{-tx}) \
&=& ; E(e^{t(-x)}) \
&=& ; M_{-X}(t) \
&=& ; M_X(t)qquadqquadtext{since $X$ and $-X$ are identically distributed.}
end{eqnarray*}
answered Sep 27 '15 at 17:13
Mick AMick A
8,8502825
$endgroup$
It looks like you've got the first part OK. For the second part you need to show that $M_X(-t)=M_X(t)$, which you can do by making use of the first part:
begin{eqnarray*}
M_X(-t) &=& ; E(e^{-tx}) \
&=& ; E(e^{t(-x)}) \
&=& ; M_{-X}(t) \
&=& ; M_X(t)qquadqquadtext{since $X$ and $-X$ are identically distributed.}
end{eqnarray*}
answered Sep 27 '15 at 17:13
Mick AMick A
8,8502825
answered Sep 27 '15 at 17:13
Mick AMick A
8,8502825
answered Sep 27 '15 at 17:13
Mick AMick A
8,8502825
answered Sep 27 '15 at 17:13
Mick AMick A
8,8502825
8,8502825
add a comment |
add a comment |
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$begingroup$
you mean $f_X(x) = f_X(-x)$
$endgroup$
– Ant
Sep 27 '15 at 17:16