Cumulative function of exponential bilateral function












1














I'm trying to calculate the cumulative function of a Laplace distribution whose PDF is




$f_X(x)=frac{lambda}{2}e^{-lambda|x|},forall x in mathbb{R}$




I proved that, if $x<0$:



$int_{-infty}^{x}frac{lambda}{2}e^{-lambda|x|}dx=[frac{lambda}{2}e^{lambda x}]_{-infty}^{x} =frac{lambda}{2}e^{lambda x}$



But I have difficulties in the other case. I obtained, if $xgeq0$:



$mathbb{P}(Xgeq 0)=mathbb{P}(X=0 bigcup X>0)=mathbb{P}(X=0)+mathbb{P}(X>0)=frac{lambda}{2}+int_{0}^{x}frac{lambda}{2}e^{-lambda|x|}dx=frac{lambda}{2}+[frac{lambda}{2}e^{-lambda x}]_{0}^{x}=frac{lambda}{2}e^{-lambda x}$



but I should have gotten $1-frac{lambda}{2}e^{-lambda x}$.
Where i wrong?



Thanks for any help!






probability integration probability-distributions density-function







share|cite|improve this question






















  • How are you integrating the $frac{lambda}{2}mathrm{e}^{lambda x}$?
    – Chinny84
    Dec 4 '18 at 15:36










  • @Chinny84 Sorry, i didn't understand the question.
    – Marco Pittella
    Dec 4 '18 at 15:40










  • For a continuous distribution, $mathbb{P}left(X = 0right) neq f_{X}left(xright)$. You should treat $mathbb{P}left(X = 0right) = 0$.
    – rzch
    Dec 4 '18 at 15:40












  • @MarcoPittella the integral should be divided by $lambda$. But you should see below.
    – Chinny84
    Dec 4 '18 at 16:06










  • I don't understand why. Maybe because, from the exponential distribution, $int_{0}^{+infty}lambda e^{-lambda x}dx=1Rightarrow int_{0}^{+infty} e^{-lambda x}dx=frac{1}{lambda}$ ?
    – Marco Pittella
    Dec 4 '18 at 16:43
















1














I'm trying to calculate the cumulative function of a Laplace distribution whose PDF is




$f_X(x)=frac{lambda}{2}e^{-lambda|x|},forall x in mathbb{R}$




I proved that, if $x<0$:



$int_{-infty}^{x}frac{lambda}{2}e^{-lambda|x|}dx=[frac{lambda}{2}e^{lambda x}]_{-infty}^{x} =frac{lambda}{2}e^{lambda x}$



But I have difficulties in the other case. I obtained, if $xgeq0$:



$mathbb{P}(Xgeq 0)=mathbb{P}(X=0 bigcup X>0)=mathbb{P}(X=0)+mathbb{P}(X>0)=frac{lambda}{2}+int_{0}^{x}frac{lambda}{2}e^{-lambda|x|}dx=frac{lambda}{2}+[frac{lambda}{2}e^{-lambda x}]_{0}^{x}=frac{lambda}{2}e^{-lambda x}$



but I should have gotten $1-frac{lambda}{2}e^{-lambda x}$.
Where i wrong?



Thanks for any help!






probability integration probability-distributions density-function







share|cite|improve this question






















  • How are you integrating the $frac{lambda}{2}mathrm{e}^{lambda x}$?
    – Chinny84
    Dec 4 '18 at 15:36










  • @Chinny84 Sorry, i didn't understand the question.
    – Marco Pittella
    Dec 4 '18 at 15:40










  • For a continuous distribution, $mathbb{P}left(X = 0right) neq f_{X}left(xright)$. You should treat $mathbb{P}left(X = 0right) = 0$.
    – rzch
    Dec 4 '18 at 15:40












  • @MarcoPittella the integral should be divided by $lambda$. But you should see below.
    – Chinny84
    Dec 4 '18 at 16:06










  • I don't understand why. Maybe because, from the exponential distribution, $int_{0}^{+infty}lambda e^{-lambda x}dx=1Rightarrow int_{0}^{+infty} e^{-lambda x}dx=frac{1}{lambda}$ ?
    – Marco Pittella
    Dec 4 '18 at 16:43














1












1








1







I'm trying to calculate the cumulative function of a Laplace distribution whose PDF is




$f_X(x)=frac{lambda}{2}e^{-lambda|x|},forall x in mathbb{R}$




I proved that, if $x<0$:



$int_{-infty}^{x}frac{lambda}{2}e^{-lambda|x|}dx=[frac{lambda}{2}e^{lambda x}]_{-infty}^{x} =frac{lambda}{2}e^{lambda x}$



But I have difficulties in the other case. I obtained, if $xgeq0$:



$mathbb{P}(Xgeq 0)=mathbb{P}(X=0 bigcup X>0)=mathbb{P}(X=0)+mathbb{P}(X>0)=frac{lambda}{2}+int_{0}^{x}frac{lambda}{2}e^{-lambda|x|}dx=frac{lambda}{2}+[frac{lambda}{2}e^{-lambda x}]_{0}^{x}=frac{lambda}{2}e^{-lambda x}$



but I should have gotten $1-frac{lambda}{2}e^{-lambda x}$.
Where i wrong?



Thanks for any help!






probability integration probability-distributions density-function







share|cite|improve this question













I'm trying to calculate the cumulative function of a Laplace distribution whose PDF is




$f_X(x)=frac{lambda}{2}e^{-lambda|x|},forall x in mathbb{R}$




I proved that, if $x<0$:



$int_{-infty}^{x}frac{lambda}{2}e^{-lambda|x|}dx=[frac{lambda}{2}e^{lambda x}]_{-infty}^{x} =frac{lambda}{2}e^{lambda x}$



But I have difficulties in the other case. I obtained, if $xgeq0$:



$mathbb{P}(Xgeq 0)=mathbb{P}(X=0 bigcup X>0)=mathbb{P}(X=0)+mathbb{P}(X>0)=frac{lambda}{2}+int_{0}^{x}frac{lambda}{2}e^{-lambda|x|}dx=frac{lambda}{2}+[frac{lambda}{2}e^{-lambda x}]_{0}^{x}=frac{lambda}{2}e^{-lambda x}$



but I should have gotten $1-frac{lambda}{2}e^{-lambda x}$.
Where i wrong?



Thanks for any help!






probability integration probability-distributions density-function




probability integration probability-distributions density-function






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 4 '18 at 15:30









Marco Pittella

1258




1258












  • How are you integrating the $frac{lambda}{2}mathrm{e}^{lambda x}$?
    – Chinny84
    Dec 4 '18 at 15:36










  • @Chinny84 Sorry, i didn't understand the question.
    – Marco Pittella
    Dec 4 '18 at 15:40










  • For a continuous distribution, $mathbb{P}left(X = 0right) neq f_{X}left(xright)$. You should treat $mathbb{P}left(X = 0right) = 0$.
    – rzch
    Dec 4 '18 at 15:40












  • @MarcoPittella the integral should be divided by $lambda$. But you should see below.
    – Chinny84
    Dec 4 '18 at 16:06










  • I don't understand why. Maybe because, from the exponential distribution, $int_{0}^{+infty}lambda e^{-lambda x}dx=1Rightarrow int_{0}^{+infty} e^{-lambda x}dx=frac{1}{lambda}$ ?
    – Marco Pittella
    Dec 4 '18 at 16:43


















  • How are you integrating the $frac{lambda}{2}mathrm{e}^{lambda x}$?
    – Chinny84
    Dec 4 '18 at 15:36










  • @Chinny84 Sorry, i didn't understand the question.
    – Marco Pittella
    Dec 4 '18 at 15:40










  • For a continuous distribution, $mathbb{P}left(X = 0right) neq f_{X}left(xright)$. You should treat $mathbb{P}left(X = 0right) = 0$.
    – rzch
    Dec 4 '18 at 15:40












  • @MarcoPittella the integral should be divided by $lambda$. But you should see below.
    – Chinny84
    Dec 4 '18 at 16:06










  • I don't understand why. Maybe because, from the exponential distribution, $int_{0}^{+infty}lambda e^{-lambda x}dx=1Rightarrow int_{0}^{+infty} e^{-lambda x}dx=frac{1}{lambda}$ ?
    – Marco Pittella
    Dec 4 '18 at 16:43
















How are you integrating the $frac{lambda}{2}mathrm{e}^{lambda x}$?
– Chinny84
Dec 4 '18 at 15:36




How are you integrating the $frac{lambda}{2}mathrm{e}^{lambda x}$?
– Chinny84
Dec 4 '18 at 15:36












@Chinny84 Sorry, i didn't understand the question.
– Marco Pittella
Dec 4 '18 at 15:40




@Chinny84 Sorry, i didn't understand the question.
– Marco Pittella
Dec 4 '18 at 15:40












For a continuous distribution, $mathbb{P}left(X = 0right) neq f_{X}left(xright)$. You should treat $mathbb{P}left(X = 0right) = 0$.
– rzch
Dec 4 '18 at 15:40






For a continuous distribution, $mathbb{P}left(X = 0right) neq f_{X}left(xright)$. You should treat $mathbb{P}left(X = 0right) = 0$.
– rzch
Dec 4 '18 at 15:40














@MarcoPittella the integral should be divided by $lambda$. But you should see below.
– Chinny84
Dec 4 '18 at 16:06




@MarcoPittella the integral should be divided by $lambda$. But you should see below.
– Chinny84
Dec 4 '18 at 16:06












I don't understand why. Maybe because, from the exponential distribution, $int_{0}^{+infty}lambda e^{-lambda x}dx=1Rightarrow int_{0}^{+infty} e^{-lambda x}dx=frac{1}{lambda}$ ?
– Marco Pittella
Dec 4 '18 at 16:43




I don't understand why. Maybe because, from the exponential distribution, $int_{0}^{+infty}lambda e^{-lambda x}dx=1Rightarrow int_{0}^{+infty} e^{-lambda x}dx=frac{1}{lambda}$ ?
– Marco Pittella
Dec 4 '18 at 16:43










1 Answer
1






active

oldest

votes


















1














For $x < 0$,



$$
int_{-infty}^{x}{rm d}t~ frac{lambda}{2}e^{-lambda |t|} = frac{e^{lambda x}}{2} tag{1}
$$



For $x ge 0$



begin{eqnarray}
int_{-infty}^{x}{rm d}t ~ frac{lambda}{2}e^{-lambda |t|} &=& color{blue}{int_{-infty}^{0}{rm d}t ~ frac{lambda}{2}e^{-lambda |t|}} + color{red}{int_{0}^{x}{rm d}t ~ frac{lambda}{2}e^{-lambda |t|}} \
&=& color{blue}{frac{1}{2}e^{lambdacdot 0}} + color{red}{frac{1}{2} - frac{1}{2}e^{-lambda x}} \
&=& 1 - frac{1}{2}e^{-lambda x} tag{2}
end{eqnarray}






share|cite|improve this answer





















  • Thanks for your answer, but i don't understand all passages. Could you made them more clear?
    – Marco Pittella
    Dec 4 '18 at 16:45










  • @MarcoPittella Sure, which part is unclear?
    – caverac
    Dec 4 '18 at 16:46










  • Above it was said of dividing by $lambda$. In fact i don't understand why you eliminate, for $x<0$, the parameter from the numerator (not for nothing i wrote $frac{lambda}{2}e^{lambda x}$). Then, i don't understand why for $xgeq0$ you start from $-infty$. The half-line of negative reals should not be excluded for hypothesis?
    – Marco Pittella
    Dec 4 '18 at 16:54












  • @MarcoPittella I see, you are confused with the integral $$ int e^{color{red}{a} x}{rm d}x = frac{1}{color{red}{a}} e^{ax} $$
    – caverac
    Dec 4 '18 at 17:01










  • Right, now I understood. Could you also explain me why, for $xgeq0$, you add the integral in blue? Aren't we trying to calculate for which probability the $x$ is positive?
    – Marco Pittella
    Dec 4 '18 at 17:38











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














For $x < 0$,



$$
int_{-infty}^{x}{rm d}t~ frac{lambda}{2}e^{-lambda |t|} = frac{e^{lambda x}}{2} tag{1}
$$



For $x ge 0$



begin{eqnarray}
int_{-infty}^{x}{rm d}t ~ frac{lambda}{2}e^{-lambda |t|} &=& color{blue}{int_{-infty}^{0}{rm d}t ~ frac{lambda}{2}e^{-lambda |t|}} + color{red}{int_{0}^{x}{rm d}t ~ frac{lambda}{2}e^{-lambda |t|}} \
&=& color{blue}{frac{1}{2}e^{lambdacdot 0}} + color{red}{frac{1}{2} - frac{1}{2}e^{-lambda x}} \
&=& 1 - frac{1}{2}e^{-lambda x} tag{2}
end{eqnarray}






share|cite|improve this answer





















  • Thanks for your answer, but i don't understand all passages. Could you made them more clear?
    – Marco Pittella
    Dec 4 '18 at 16:45










  • @MarcoPittella Sure, which part is unclear?
    – caverac
    Dec 4 '18 at 16:46










  • Above it was said of dividing by $lambda$. In fact i don't understand why you eliminate, for $x<0$, the parameter from the numerator (not for nothing i wrote $frac{lambda}{2}e^{lambda x}$). Then, i don't understand why for $xgeq0$ you start from $-infty$. The half-line of negative reals should not be excluded for hypothesis?
    – Marco Pittella
    Dec 4 '18 at 16:54












  • @MarcoPittella I see, you are confused with the integral $$ int e^{color{red}{a} x}{rm d}x = frac{1}{color{red}{a}} e^{ax} $$
    – caverac
    Dec 4 '18 at 17:01










  • Right, now I understood. Could you also explain me why, for $xgeq0$, you add the integral in blue? Aren't we trying to calculate for which probability the $x$ is positive?
    – Marco Pittella
    Dec 4 '18 at 17:38
















1














For $x < 0$,



$$
int_{-infty}^{x}{rm d}t~ frac{lambda}{2}e^{-lambda |t|} = frac{e^{lambda x}}{2} tag{1}
$$



For $x ge 0$



begin{eqnarray}
int_{-infty}^{x}{rm d}t ~ frac{lambda}{2}e^{-lambda |t|} &=& color{blue}{int_{-infty}^{0}{rm d}t ~ frac{lambda}{2}e^{-lambda |t|}} + color{red}{int_{0}^{x}{rm d}t ~ frac{lambda}{2}e^{-lambda |t|}} \
&=& color{blue}{frac{1}{2}e^{lambdacdot 0}} + color{red}{frac{1}{2} - frac{1}{2}e^{-lambda x}} \
&=& 1 - frac{1}{2}e^{-lambda x} tag{2}
end{eqnarray}






share|cite|improve this answer





















  • Thanks for your answer, but i don't understand all passages. Could you made them more clear?
    – Marco Pittella
    Dec 4 '18 at 16:45










  • @MarcoPittella Sure, which part is unclear?
    – caverac
    Dec 4 '18 at 16:46










  • Above it was said of dividing by $lambda$. In fact i don't understand why you eliminate, for $x<0$, the parameter from the numerator (not for nothing i wrote $frac{lambda}{2}e^{lambda x}$). Then, i don't understand why for $xgeq0$ you start from $-infty$. The half-line of negative reals should not be excluded for hypothesis?
    – Marco Pittella
    Dec 4 '18 at 16:54












  • @MarcoPittella I see, you are confused with the integral $$ int e^{color{red}{a} x}{rm d}x = frac{1}{color{red}{a}} e^{ax} $$
    – caverac
    Dec 4 '18 at 17:01










  • Right, now I understood. Could you also explain me why, for $xgeq0$, you add the integral in blue? Aren't we trying to calculate for which probability the $x$ is positive?
    – Marco Pittella
    Dec 4 '18 at 17:38














1












1








1






For $x < 0$,



$$
int_{-infty}^{x}{rm d}t~ frac{lambda}{2}e^{-lambda |t|} = frac{e^{lambda x}}{2} tag{1}
$$



For $x ge 0$



begin{eqnarray}
int_{-infty}^{x}{rm d}t ~ frac{lambda}{2}e^{-lambda |t|} &=& color{blue}{int_{-infty}^{0}{rm d}t ~ frac{lambda}{2}e^{-lambda |t|}} + color{red}{int_{0}^{x}{rm d}t ~ frac{lambda}{2}e^{-lambda |t|}} \
&=& color{blue}{frac{1}{2}e^{lambdacdot 0}} + color{red}{frac{1}{2} - frac{1}{2}e^{-lambda x}} \
&=& 1 - frac{1}{2}e^{-lambda x} tag{2}
end{eqnarray}






share|cite|improve this answer












For $x < 0$,



$$
int_{-infty}^{x}{rm d}t~ frac{lambda}{2}e^{-lambda |t|} = frac{e^{lambda x}}{2} tag{1}
$$



For $x ge 0$



begin{eqnarray}
int_{-infty}^{x}{rm d}t ~ frac{lambda}{2}e^{-lambda |t|} &=& color{blue}{int_{-infty}^{0}{rm d}t ~ frac{lambda}{2}e^{-lambda |t|}} + color{red}{int_{0}^{x}{rm d}t ~ frac{lambda}{2}e^{-lambda |t|}} \
&=& color{blue}{frac{1}{2}e^{lambdacdot 0}} + color{red}{frac{1}{2} - frac{1}{2}e^{-lambda x}} \
&=& 1 - frac{1}{2}e^{-lambda x} tag{2}
end{eqnarray}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 15:45









caverac

13.9k21130




13.9k21130












  • Thanks for your answer, but i don't understand all passages. Could you made them more clear?
    – Marco Pittella
    Dec 4 '18 at 16:45










  • @MarcoPittella Sure, which part is unclear?
    – caverac
    Dec 4 '18 at 16:46










  • Above it was said of dividing by $lambda$. In fact i don't understand why you eliminate, for $x<0$, the parameter from the numerator (not for nothing i wrote $frac{lambda}{2}e^{lambda x}$). Then, i don't understand why for $xgeq0$ you start from $-infty$. The half-line of negative reals should not be excluded for hypothesis?
    – Marco Pittella
    Dec 4 '18 at 16:54












  • @MarcoPittella I see, you are confused with the integral $$ int e^{color{red}{a} x}{rm d}x = frac{1}{color{red}{a}} e^{ax} $$
    – caverac
    Dec 4 '18 at 17:01










  • Right, now I understood. Could you also explain me why, for $xgeq0$, you add the integral in blue? Aren't we trying to calculate for which probability the $x$ is positive?
    – Marco Pittella
    Dec 4 '18 at 17:38


















  • Thanks for your answer, but i don't understand all passages. Could you made them more clear?
    – Marco Pittella
    Dec 4 '18 at 16:45










  • @MarcoPittella Sure, which part is unclear?
    – caverac
    Dec 4 '18 at 16:46










  • Above it was said of dividing by $lambda$. In fact i don't understand why you eliminate, for $x<0$, the parameter from the numerator (not for nothing i wrote $frac{lambda}{2}e^{lambda x}$). Then, i don't understand why for $xgeq0$ you start from $-infty$. The half-line of negative reals should not be excluded for hypothesis?
    – Marco Pittella
    Dec 4 '18 at 16:54












  • @MarcoPittella I see, you are confused with the integral $$ int e^{color{red}{a} x}{rm d}x = frac{1}{color{red}{a}} e^{ax} $$
    – caverac
    Dec 4 '18 at 17:01










  • Right, now I understood. Could you also explain me why, for $xgeq0$, you add the integral in blue? Aren't we trying to calculate for which probability the $x$ is positive?
    – Marco Pittella
    Dec 4 '18 at 17:38
















Thanks for your answer, but i don't understand all passages. Could you made them more clear?
– Marco Pittella
Dec 4 '18 at 16:45




Thanks for your answer, but i don't understand all passages. Could you made them more clear?
– Marco Pittella
Dec 4 '18 at 16:45












@MarcoPittella Sure, which part is unclear?
– caverac
Dec 4 '18 at 16:46




@MarcoPittella Sure, which part is unclear?
– caverac
Dec 4 '18 at 16:46












Above it was said of dividing by $lambda$. In fact i don't understand why you eliminate, for $x<0$, the parameter from the numerator (not for nothing i wrote $frac{lambda}{2}e^{lambda x}$). Then, i don't understand why for $xgeq0$ you start from $-infty$. The half-line of negative reals should not be excluded for hypothesis?
– Marco Pittella
Dec 4 '18 at 16:54






Above it was said of dividing by $lambda$. In fact i don't understand why you eliminate, for $x<0$, the parameter from the numerator (not for nothing i wrote $frac{lambda}{2}e^{lambda x}$). Then, i don't understand why for $xgeq0$ you start from $-infty$. The half-line of negative reals should not be excluded for hypothesis?
– Marco Pittella
Dec 4 '18 at 16:54














@MarcoPittella I see, you are confused with the integral $$ int e^{color{red}{a} x}{rm d}x = frac{1}{color{red}{a}} e^{ax} $$
– caverac
Dec 4 '18 at 17:01




@MarcoPittella I see, you are confused with the integral $$ int e^{color{red}{a} x}{rm d}x = frac{1}{color{red}{a}} e^{ax} $$
– caverac
Dec 4 '18 at 17:01












Right, now I understood. Could you also explain me why, for $xgeq0$, you add the integral in blue? Aren't we trying to calculate for which probability the $x$ is positive?
– Marco Pittella
Dec 4 '18 at 17:38




Right, now I understood. Could you also explain me why, for $xgeq0$, you add the integral in blue? Aren't we trying to calculate for which probability the $x$ is positive?
– Marco Pittella
Dec 4 '18 at 17:38


















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Dijon

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Dijon.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} De haut en bas, de gauche à droite : Panorama du centre historique vu depuis la Cathédrale Saint-Bénigne, le tramway passant sous l'Auditorium , la place François Rude, les Tuiles vernissées de l'Hôtel Aubriot, le centre commercial Toison d'or et le jardin Darcy. Blason Logo Administration Pays   France Région Bourgogne-Franche-Comté (préfecture) Département Côte-d'Or (préfecture) Arrondissement Dijon (chef-lieu) Canton bureau centralisateur de 6 cantons : Dijon-1, 2, 3, 4, 5, 6 Intercommunalité Dijon Métropole Maire Mandat François Rebsamen 2015-2020 Code postal 21000 (comportait aussi le code 21100 quand la ville possédait deux bureaux distributeurs) Code commune 21231 Démographie Gentilé ...
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Guerrita

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Guerrita Présentation Nom de naissance Rafael Guerra Bejarano Apodo Guerrita Naissance 6 mars 1862 Cordoue Décès 21 février 1941 (à 78 ans) Nationalité Espagnol Carrière Alternative 29 septembre 1887 à Madrid Parrain, Lagartijo Fin de carrière 15 octobre 1899 Distinction Nommé « Deuxième Calife de la Tauromachie » modifier   Rafael Guerra Bejarano dit « Guerrita » , né à Cordoue (Espagne) le 6 mars 1862, mort à Cordoue le 21 février 1941, était un matador espagnol. Présentation | « Guerrita » fut sans doute l’un des plus grands matadors de son époque, peut-être même le plus grand : insurpassable avec les banderilles, d’une parfaite maîtrise avec la muleta , infaillible avec l’épée. Il se fait d'abord banderillero comme il est fréquent à l'époque et acquiert vite une réputation qui dépasse celle du maestro qu’il assiste, d’abord « Bocanegra », puis Fernando Gómez « El Gall...
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