Mixing
Finding characteristic function then density function is given
$begingroup$
Random variable $xi$ is distributed by symmetrical principle with density function $frac{1}{2a} mathcal{1}_{[-a,a]}(x),$ here $a>0$. I need to find characteristic function.
I never seen anything like this before. I tried to find some theory about it. And I found that $f(t)=int_{-infty}^{+infty}e^{itx} dF(X)$, where $f(t)$ is characteristic function and $F(x)$ is distribution function. Also I found that density function is $p(x)=frac{1}{2pi}int_{-infty}^{+infty}e^{-itx} f(t)dt.$ Do I can do something with this? How could I find characteristic function?
probability probability-theory density-function characteristic-functions
asked Jan 24 at 12:51
AtstovasAtstovas
1139
$endgroup$
add a comment |
$begingroup$
Random variable $xi$ is distributed by symmetrical principle with density function $frac{1}{2a} mathcal{1}_{[-a,a]}(x),$ here $a>0$. I need to find characteristic function.
I never seen anything like this before. I tried to find some theory about it. And I found that $f(t)=int_{-infty}^{+infty}e^{itx} dF(X)$, where $f(t)$ is characteristic function and $F(x)$ is distribution function. Also I found that density function is $p(x)=frac{1}{2pi}int_{-infty}^{+infty}e^{-itx} f(t)dt.$ Do I can do something with this? How could I find characteristic function?
probability probability-theory density-function characteristic-functions
asked Jan 24 at 12:51
AtstovasAtstovas
1139
$endgroup$
add a comment |
$begingroup$
Random variable $xi$ is distributed by symmetrical principle with density function $frac{1}{2a} mathcal{1}_{[-a,a]}(x),$ here $a>0$. I need to find characteristic function.
I never seen anything like this before. I tried to find some theory about it. And I found that $f(t)=int_{-infty}^{+infty}e^{itx} dF(X)$, where $f(t)$ is characteristic function and $F(x)$ is distribution function. Also I found that density function is $p(x)=frac{1}{2pi}int_{-infty}^{+infty}e^{-itx} f(t)dt.$ Do I can do something with this? How could I find characteristic function?
probability probability-theory density-function characteristic-functions
asked Jan 24 at 12:51
AtstovasAtstovas
1139
$endgroup$
Random variable $xi$ is distributed by symmetrical principle with density function $frac{1}{2a} mathcal{1}_{[-a,a]}(x),$ here $a>0$. I need to find characteristic function.
I never seen anything like this before. I tried to find some theory about it. And I found that $f(t)=int_{-infty}^{+infty}e^{itx} dF(X)$, where $f(t)$ is characteristic function and $F(x)$ is distribution function. Also I found that density function is $p(x)=frac{1}{2pi}int_{-infty}^{+infty}e^{-itx} f(t)dt.$ Do I can do something with this? How could I find characteristic function?
probability probability-theory density-function characteristic-functions
probability probability-theory density-function characteristic-functions
asked Jan 24 at 12:51
AtstovasAtstovas
1139
asked Jan 24 at 12:51
AtstovasAtstovas
1139
asked Jan 24 at 12:51
AtstovasAtstovas
1139
asked Jan 24 at 12:51
AtstovasAtstovas
1139
asked Jan 24 at 12:51
AtstovasAtstovas
1139
1139
add a comment |
add a comment |
1 Answer
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For continuous distributions $dF=pdx$ on the support, in this case $[-a,,a]$, so $$f(t)=int_{-a}^afrac{1}{2a}exp itx dx=left[frac{exp itx}{2ita}right]_{-a}^a=frac{1}{ta}sin ta$$(or $1$ at $t=0$ by continuity).
answered Jan 24 at 13:05
J.G.J.G.
30.3k23148
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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$begingroup$
For continuous distributions $dF=pdx$ on the support, in this case $[-a,,a]$, so $$f(t)=int_{-a}^afrac{1}{2a}exp itx dx=left[frac{exp itx}{2ita}right]_{-a}^a=frac{1}{ta}sin ta$$(or $1$ at $t=0$ by continuity).
answered Jan 24 at 13:05
J.G.J.G.
30.3k23148
$endgroup$
add a comment |
$begingroup$
For continuous distributions $dF=pdx$ on the support, in this case $[-a,,a]$, so $$f(t)=int_{-a}^afrac{1}{2a}exp itx dx=left[frac{exp itx}{2ita}right]_{-a}^a=frac{1}{ta}sin ta$$(or $1$ at $t=0$ by continuity).
answered Jan 24 at 13:05
J.G.J.G.
30.3k23148
$endgroup$
add a comment |
$begingroup$
For continuous distributions $dF=pdx$ on the support, in this case $[-a,,a]$, so $$f(t)=int_{-a}^afrac{1}{2a}exp itx dx=left[frac{exp itx}{2ita}right]_{-a}^a=frac{1}{ta}sin ta$$(or $1$ at $t=0$ by continuity).
answered Jan 24 at 13:05
J.G.J.G.
30.3k23148
$endgroup$
For continuous distributions $dF=pdx$ on the support, in this case $[-a,,a]$, so $$f(t)=int_{-a}^afrac{1}{2a}exp itx dx=left[frac{exp itx}{2ita}right]_{-a}^a=frac{1}{ta}sin ta$$(or $1$ at $t=0$ by continuity).
answered Jan 24 at 13:05
J.G.J.G.
30.3k23148
answered Jan 24 at 13:05
J.G.J.G.
30.3k23148
answered Jan 24 at 13:05
J.G.J.G.
30.3k23148
answered Jan 24 at 13:05
J.G.J.G.
30.3k23148
30.3k23148
add a comment |
add a comment |
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