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Given that $X$ is a random variable with cdf
$F(x)= 1-e^{-x}$ when $x>0$ and $0$ when $x le 0$, find $P(0 le e^X le 4)$.



I know the formula for $P(a le X le b)$ but here how I convert $e^X$ into $X$ . I assume that $y=e^X$ but in that case if I take log on both sides then it is not defined at $X=0$.






probability probability-distributions







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    up vote
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    down vote

    favorite












    Given that $X$ is a random variable with cdf
    $F(x)= 1-e^{-x}$ when $x>0$ and $0$ when $x le 0$, find $P(0 le e^X le 4)$.



    I know the formula for $P(a le X le b)$ but here how I convert $e^X$ into $X$ . I assume that $y=e^X$ but in that case if I take log on both sides then it is not defined at $X=0$.






    probability probability-distributions







    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Given that $X$ is a random variable with cdf
      $F(x)= 1-e^{-x}$ when $x>0$ and $0$ when $x le 0$, find $P(0 le e^X le 4)$.



      I know the formula for $P(a le X le b)$ but here how I convert $e^X$ into $X$ . I assume that $y=e^X$ but in that case if I take log on both sides then it is not defined at $X=0$.






      probability probability-distributions







      share|cite|improve this question















      Given that $X$ is a random variable with cdf
      $F(x)= 1-e^{-x}$ when $x>0$ and $0$ when $x le 0$, find $P(0 le e^X le 4)$.



      I know the formula for $P(a le X le b)$ but here how I convert $e^X$ into $X$ . I assume that $y=e^X$ but in that case if I take log on both sides then it is not defined at $X=0$.






      probability probability-distributions




      probability probability-distributions






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 22 at 18:26









      Daniel

      1,516210




      1,516210










      asked Nov 22 at 17:56









      tiger

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          Since $exp(x)>0$ for any $x in mathbb{R}$, we have
          $$P(0 le e^X le 4) = P(0<e^X le 4).$$
          Moreover, we know that $log$ maps $(0,4]$ injective onto $(-infty,log(4)]$, because this function is monotonically increasing and continuous. Thus
          $$P(0<e^X le 4) = P(X le log(4)) = F(log(4)) = 1-e^{-log(4)}=1-1/4=3/4.$$






          share|cite|improve this answer





















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            Since $exp(x)>0$ for any $x in mathbb{R}$, we have
            $$P(0 le e^X le 4) = P(0<e^X le 4).$$
            Moreover, we know that $log$ maps $(0,4]$ injective onto $(-infty,log(4)]$, because this function is monotonically increasing and continuous. Thus
            $$P(0<e^X le 4) = P(X le log(4)) = F(log(4)) = 1-e^{-log(4)}=1-1/4=3/4.$$






            share|cite|improve this answer

























              up vote
              1
              down vote













              Since $exp(x)>0$ for any $x in mathbb{R}$, we have
              $$P(0 le e^X le 4) = P(0<e^X le 4).$$
              Moreover, we know that $log$ maps $(0,4]$ injective onto $(-infty,log(4)]$, because this function is monotonically increasing and continuous. Thus
              $$P(0<e^X le 4) = P(X le log(4)) = F(log(4)) = 1-e^{-log(4)}=1-1/4=3/4.$$






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                Since $exp(x)>0$ for any $x in mathbb{R}$, we have
                $$P(0 le e^X le 4) = P(0<e^X le 4).$$
                Moreover, we know that $log$ maps $(0,4]$ injective onto $(-infty,log(4)]$, because this function is monotonically increasing and continuous. Thus
                $$P(0<e^X le 4) = P(X le log(4)) = F(log(4)) = 1-e^{-log(4)}=1-1/4=3/4.$$






                share|cite|improve this answer












                Since $exp(x)>0$ for any $x in mathbb{R}$, we have
                $$P(0 le e^X le 4) = P(0<e^X le 4).$$
                Moreover, we know that $log$ maps $(0,4]$ injective onto $(-infty,log(4)]$, because this function is monotonically increasing and continuous. Thus
                $$P(0<e^X le 4) = P(X le log(4)) = F(log(4)) = 1-e^{-log(4)}=1-1/4=3/4.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 22 at 18:03









                p4sch

                4,800217




                4,800217






























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