Finding Cumulative distribution function of $f_x(X) = frac14 cdot |X|$Cumulative distribution function and expected valueFinding Cumulative Distribution Function given two independent pdfsFind continuous stochastic variable $X$ with PDF $f_X = frac1x^2$Cumulative distribution function of sub of two random variableFinding the probability density from cumulative distribution functioncumulative distribution function method transformationFinding density function given a cumulative distributionCumulative Distribution Function for a Certain rangeCummulative Distribution Function: Sum of two independent exp-distributed random variablesCumulative distribution function to probability density
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Finding Cumulative distribution function of $f_x(X) = frac14 cdot |X|$
Cumulative distribution function and expected valueFinding Cumulative Distribution Function given two independent pdfsFind continuous stochastic variable $X$ with PDF $f_X = frac1x^2$Cumulative distribution function of sub of two random variableFinding the probability density from cumulative distribution functioncumulative distribution function method transformationFinding density function given a cumulative distributionCumulative Distribution Function for a Certain rangeCummulative Distribution Function: Sum of two independent exp-distributed random variablesCumulative distribution function to probability density
$begingroup$
Let $f_X(x) = frac14cdot |X|$ a probability density function.
$-2<X<2$;
Find the Cumulative distribution function of $f_X(x)$.
What I thought was the answer is $F_X = frac12 + fracx8$. Since a CDF is a function which tells us the probability of a $P(X<a)$, given certain.
Drawing the PDF generats a graph of a symetric function, which gives us a probability of a least $frac12$.
However I found out it is $F_x = PleftXle xright=frac12-frac12cdot |x|cdot |fracx4|= frac12 - fracx^28$, for $-2 < x < 0$.
I can't understand why.
probability probability-theory density-function
asked 2 days ago
AlanAlan
1,3991021
$endgroup$
add a comment |
$begingroup$
Let $f_X(x) = frac14cdot |X|$ a probability density function.
$-2<X<2$;
Find the Cumulative distribution function of $f_X(x)$.
What I thought was the answer is $F_X = frac12 + fracx8$. Since a CDF is a function which tells us the probability of a $P(X<a)$, given certain.
Drawing the PDF generats a graph of a symetric function, which gives us a probability of a least $frac12$.
However I found out it is $F_x = PleftXle xright=frac12-frac12cdot |x|cdot |fracx4|= frac12 - fracx^28$, for $-2 < x < 0$.
I can't understand why.
probability probability-theory density-function
asked 2 days ago
AlanAlan
1,3991021
$endgroup$
1
$begingroup$
We have $P(Xle x) =int_-2^x frac14|t|, dt $ if $xin (-2,2)$. Did you try using this? This probability is only at least $1/2$ if $xge 0$. To compute this integral, you can try doing it by cases of $x < 0$ and $x ge 0$ (for $xin (-2,2)$).
$endgroup$
– Minus One-Twelfth
2 days ago
1
$begingroup$
Do you mean it is $frac12 + fracx^28$ for $0 < x < 2$?
$endgroup$
– Infiaria
2 days ago
$begingroup$
@Infiaria Yes, it is. Thank you.
$endgroup$
– Alan
2 days ago
$begingroup$
@MinusOne-Twelfth Got it to work, thanks.
$endgroup$
– Alan
2 days ago
add a comment |
$begingroup$
Let $f_X(x) = frac14cdot |X|$ a probability density function.
$-2<X<2$;
Find the Cumulative distribution function of $f_X(x)$.
What I thought was the answer is $F_X = frac12 + fracx8$. Since a CDF is a function which tells us the probability of a $P(X<a)$, given certain.
Drawing the PDF generats a graph of a symetric function, which gives us a probability of a least $frac12$.
However I found out it is $F_x = PleftXle xright=frac12-frac12cdot |x|cdot |fracx4|= frac12 - fracx^28$, for $-2 < x < 0$.
I can't understand why.
probability probability-theory density-function
asked 2 days ago
AlanAlan
1,3991021
$endgroup$
Let $f_X(x) = frac14cdot |X|$ a probability density function.
$-2<X<2$;
Find the Cumulative distribution function of $f_X(x)$.
What I thought was the answer is $F_X = frac12 + fracx8$. Since a CDF is a function which tells us the probability of a $P(X<a)$, given certain.
Drawing the PDF generats a graph of a symetric function, which gives us a probability of a least $frac12$.
However I found out it is $F_x = PleftXle xright=frac12-frac12cdot |x|cdot |fracx4|= frac12 - fracx^28$, for $-2 < x < 0$.
I can't understand why.
probability probability-theory density-function
probability probability-theory density-function
asked 2 days ago
AlanAlan
1,3991021
asked 2 days ago
AlanAlan
1,3991021
edited 2 days ago
Alan
asked 2 days ago
AlanAlan
1,3991021
asked 2 days ago
AlanAlan
1,3991021
asked 2 days ago
AlanAlan
1,3991021
1,3991021
1
$begingroup$
We have $P(Xle x) =int_-2^x frac14|t|, dt $ if $xin (-2,2)$. Did you try using this? This probability is only at least $1/2$ if $xge 0$. To compute this integral, you can try doing it by cases of $x < 0$ and $x ge 0$ (for $xin (-2,2)$).
$endgroup$
– Minus One-Twelfth
2 days ago
1
$begingroup$
Do you mean it is $frac12 + fracx^28$ for $0 < x < 2$?
$endgroup$
– Infiaria
2 days ago
$begingroup$
@Infiaria Yes, it is. Thank you.
$endgroup$
– Alan
2 days ago
$begingroup$
@MinusOne-Twelfth Got it to work, thanks.
$endgroup$
– Alan
2 days ago
add a comment |
1
$begingroup$
We have $P(Xle x) =int_-2^x frac14|t|, dt $ if $xin (-2,2)$. Did you try using this? This probability is only at least $1/2$ if $xge 0$. To compute this integral, you can try doing it by cases of $x < 0$ and $x ge 0$ (for $xin (-2,2)$).
$endgroup$
– Minus One-Twelfth
2 days ago
1
$begingroup$
Do you mean it is $frac12 + fracx^28$ for $0 < x < 2$?
$endgroup$
– Infiaria
2 days ago
$begingroup$
@Infiaria Yes, it is. Thank you.
$endgroup$
– Alan
2 days ago
$begingroup$
@MinusOne-Twelfth Got it to work, thanks.
$endgroup$
– Alan
2 days ago
1
1
$begingroup$
We have $P(Xle x) =int_-2^x frac14|t|, dt $ if $xin (-2,2)$. Did you try using this? This probability is only at least $1/2$ if $xge 0$. To compute this integral, you can try doing it by cases of $x < 0$ and $x ge 0$ (for $xin (-2,2)$).
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
We have $P(Xle x) =int_-2^x frac14|t|, dt $ if $xin (-2,2)$. Did you try using this? This probability is only at least $1/2$ if $xge 0$. To compute this integral, you can try doing it by cases of $x < 0$ and $x ge 0$ (for $xin (-2,2)$).
$endgroup$
– Minus One-Twelfth
2 days ago
1
1
$begingroup$
Do you mean it is $frac12 + fracx^28$ for $0 < x < 2$?
$endgroup$
– Infiaria
2 days ago
$begingroup$
Do you mean it is $frac12 + fracx^28$ for $0 < x < 2$?
$endgroup$
– Infiaria
2 days ago
$begingroup$
@Infiaria Yes, it is. Thank you.
$endgroup$
– Alan
2 days ago
$begingroup$
@Infiaria Yes, it is. Thank you.
$endgroup$
– Alan
2 days ago
$begingroup$
@MinusOne-Twelfth Got it to work, thanks.
$endgroup$
– Alan
2 days ago
$begingroup$
@MinusOne-Twelfth Got it to work, thanks.
$endgroup$
– Alan
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
With the help above, I made it -
I will show the answer for the interval $-2 < x < 0$. The other one is symetric.
$$int_-2^x frac14cdot |t| dt = int_-2^x frac14cdot (-t) dt $$ (since t is negative).
$$=frac14cdot int_-2^x (t) dt =- frac14cdot fract^22|_-2^x=-frac14cdot left( fracx^22 - 2 right) =frac12 - fracx^28 $$.
So at the end we get -
$$F_x = left{ beginarray*20l0&X < - 2\beginarraylfrac12 - fracx^28\frac12 + fracx^28\1endarray&beginarrayl - 2 < x < 0\0 < x < 2\x > 2endarrayendarray right.
% MathType!MTEF!2!1!+-
% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbdfgBPj
% MCPbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B
% TfMBaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbbf9v8
% qqaqFr0xc9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9
% q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaake
% aaqaaaaaaaaaWdbiaadAeapaWaaSbaaSqaa8qacaWG4baapaqabaGc
% peGaeyypa0Zaaiqaa8aabaqbaeaabiGaaaqaaiaaicdaaeaacaWGyb
% GaeyipaWJaeyOeI0IaaGOmaaabaeqabaWaaSaaaeaacaaIXaaabaGa
% aGOmaaaacqGHsisldaWcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaa
% aakeaacaaI4aaaaaqaamaalaaabaGaaGymaaqaaiaaikdaaaGaey4k
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% aaaeaacaaIXaaaaqaabeqaaiabgkHiTiaaikdacqGH8aapcaWG4bGa
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% qaaiaadIhacqGH+aGpcaaIYaaaaaaapeGaay5Eaaaaaa!5984!
$$
answered 2 days ago
AlanAlan
1,3991021
$endgroup$
add a comment |
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$begingroup$
With the help above, I made it -
I will show the answer for the interval $-2 < x < 0$. The other one is symetric.
$$int_-2^x frac14cdot |t| dt = int_-2^x frac14cdot (-t) dt $$ (since t is negative).
$$=frac14cdot int_-2^x (t) dt =- frac14cdot fract^22|_-2^x=-frac14cdot left( fracx^22 - 2 right) =frac12 - fracx^28 $$.
So at the end we get -
$$F_x = left{ beginarray*20l0&X < - 2\beginarraylfrac12 - fracx^28\frac12 + fracx^28\1endarray&beginarrayl - 2 < x < 0\0 < x < 2\x > 2endarrayendarray right.
% MathType!MTEF!2!1!+-
% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbdfgBPj
% MCPbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B
% TfMBaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbbf9v8
% qqaqFr0xc9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9
% q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaake
% aaqaaaaaaaaaWdbiaadAeapaWaaSbaaSqaa8qacaWG4baapaqabaGc
% peGaeyypa0Zaaiqaa8aabaqbaeaabiGaaaqaaiaaicdaaeaacaWGyb
% GaeyipaWJaeyOeI0IaaGOmaaabaeqabaWaaSaaaeaacaaIXaaabaGa
% aGOmaaaacqGHsisldaWcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaa
% aakeaacaaI4aaaaaqaamaalaaabaGaaGymaaqaaiaaikdaaaGaey4k
% aSYaaSaaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaaGcbaGaaGioaa
% aaaeaacaaIXaaaaqaabeqaaiabgkHiTiaaikdacqGH8aapcaWG4bGa
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% qaaiaadIhacqGH+aGpcaaIYaaaaaaapeGaay5Eaaaaaa!5984!
$$
answered 2 days ago
AlanAlan
1,3991021
$endgroup$
add a comment |
$begingroup$
With the help above, I made it -
I will show the answer for the interval $-2 < x < 0$. The other one is symetric.
$$int_-2^x frac14cdot |t| dt = int_-2^x frac14cdot (-t) dt $$ (since t is negative).
$$=frac14cdot int_-2^x (t) dt =- frac14cdot fract^22|_-2^x=-frac14cdot left( fracx^22 - 2 right) =frac12 - fracx^28 $$.
So at the end we get -
$$F_x = left{ beginarray*20l0&X < - 2\beginarraylfrac12 - fracx^28\frac12 + fracx^28\1endarray&beginarrayl - 2 < x < 0\0 < x < 2\x > 2endarrayendarray right.
% MathType!MTEF!2!1!+-
% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbdfgBPj
% MCPbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B
% TfMBaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbbf9v8
% qqaqFr0xc9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9
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% GaeyipaWJaeyOeI0IaaGOmaaabaeqabaWaaSaaaeaacaaIXaaabaGa
% aGOmaaaacqGHsisldaWcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaa
% aakeaacaaI4aaaaaqaamaalaaabaGaaGymaaqaaiaaikdaaaGaey4k
% aSYaaSaaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaaGcbaGaaGioaa
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% qaaiaadIhacqGH+aGpcaaIYaaaaaaapeGaay5Eaaaaaa!5984!
$$
answered 2 days ago
AlanAlan
1,3991021
$endgroup$
add a comment |
$begingroup$
With the help above, I made it -
I will show the answer for the interval $-2 < x < 0$. The other one is symetric.
$$int_-2^x frac14cdot |t| dt = int_-2^x frac14cdot (-t) dt $$ (since t is negative).
$$=frac14cdot int_-2^x (t) dt =- frac14cdot fract^22|_-2^x=-frac14cdot left( fracx^22 - 2 right) =frac12 - fracx^28 $$.
So at the end we get -
$$F_x = left{ beginarray*20l0&X < - 2\beginarraylfrac12 - fracx^28\frac12 + fracx^28\1endarray&beginarrayl - 2 < x < 0\0 < x < 2\x > 2endarrayendarray right.
% MathType!MTEF!2!1!+-
% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbdfgBPj
% MCPbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B
% TfMBaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbbf9v8
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% GaeyipaWJaeyOeI0IaaGOmaaabaeqabaWaaSaaaeaacaaIXaaabaGa
% aGOmaaaacqGHsisldaWcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaa
% aakeaacaaI4aaaaaqaamaalaaabaGaaGymaaqaaiaaikdaaaGaey4k
% aSYaaSaaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaaGcbaGaaGioaa
% aaaeaacaaIXaaaaqaabeqaaiabgkHiTiaaikdacqGH8aapcaWG4bGa
% eyipaWJaaGimaaqaaiaaicdacqGH8aapcaWG4bGaeyipaWJaaGOmaa
% qaaiaadIhacqGH+aGpcaaIYaaaaaaapeGaay5Eaaaaaa!5984!
$$
answered 2 days ago
AlanAlan
1,3991021
$endgroup$
With the help above, I made it -
I will show the answer for the interval $-2 < x < 0$. The other one is symetric.
$$int_-2^x frac14cdot |t| dt = int_-2^x frac14cdot (-t) dt $$ (since t is negative).
$$=frac14cdot int_-2^x (t) dt =- frac14cdot fract^22|_-2^x=-frac14cdot left( fracx^22 - 2 right) =frac12 - fracx^28 $$.
So at the end we get -
$$F_x = left{ beginarray*20l0&X < - 2\beginarraylfrac12 - fracx^28\frac12 + fracx^28\1endarray&beginarrayl - 2 < x < 0\0 < x < 2\x > 2endarrayendarray right.
% MathType!MTEF!2!1!+-
% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbdfgBPj
% MCPbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B
% TfMBaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbbf9v8
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% GaeyipaWJaeyOeI0IaaGOmaaabaeqabaWaaSaaaeaacaaIXaaabaGa
% aGOmaaaacqGHsisldaWcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaa
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% eyipaWJaaGimaaqaaiaaicdacqGH8aapcaWG4bGaeyipaWJaaGOmaa
% qaaiaadIhacqGH+aGpcaaIYaaaaaaapeGaay5Eaaaaaa!5984!
$$
answered 2 days ago
AlanAlan
1,3991021
answered 2 days ago
AlanAlan
1,3991021
answered 2 days ago
AlanAlan
1,3991021
answered 2 days ago
AlanAlan
1,3991021
1,3991021
add a comment |
add a comment |
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We have $P(Xle x) =int_-2^x frac14|t|, dt $ if $xin (-2,2)$. Did you try using this? This probability is only at least $1/2$ if $xge 0$. To compute this integral, you can try doing it by cases of $x < 0$ and $x ge 0$ (for $xin (-2,2)$).
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– Minus One-Twelfth
2 days ago
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Do you mean it is $frac12 + fracx^28$ for $0 < x < 2$?
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– Infiaria
2 days ago
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@Infiaria Yes, it is. Thank you.
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– Alan
2 days ago
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@MinusOne-Twelfth Got it to work, thanks.
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– Alan
2 days ago