Finding the moment generating function of $min(Y,1)$












0












$begingroup$


Let $Ysimtext{Exp}(1)$ be a random variable. I denote the random variable $X$ as $X=min(Y,1)$. The task is to find the moment generating function of $X$.



By simply calculating the probability I managed to find the CDF of $X$ is:
$$F_X(t) = begin{cases}
1-e^{-t}, & 0 le t <1\
1, & t ge 1\
0, & text{otherwise}
end{cases}$$



Here I got stuck. Since $X$ is not a continuous variable, it does not have a PDF, and without it I do not know how to calculate the moment generating function. ($X$ is also not discrete).



I will appreciate some help.






probability moment-generating-functions exponential-distribution







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Let $Ysimtext{Exp}(1)$ be a random variable. I denote the random variable $X$ as $X=min(Y,1)$. The task is to find the moment generating function of $X$.



    By simply calculating the probability I managed to find the CDF of $X$ is:
    $$F_X(t) = begin{cases}
    1-e^{-t}, & 0 le t <1\
    1, & t ge 1\
    0, & text{otherwise}
    end{cases}$$



    Here I got stuck. Since $X$ is not a continuous variable, it does not have a PDF, and without it I do not know how to calculate the moment generating function. ($X$ is also not discrete).



    I will appreciate some help.






    probability moment-generating-functions exponential-distribution







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $Ysimtext{Exp}(1)$ be a random variable. I denote the random variable $X$ as $X=min(Y,1)$. The task is to find the moment generating function of $X$.



      By simply calculating the probability I managed to find the CDF of $X$ is:
      $$F_X(t) = begin{cases}
      1-e^{-t}, & 0 le t <1\
      1, & t ge 1\
      0, & text{otherwise}
      end{cases}$$



      Here I got stuck. Since $X$ is not a continuous variable, it does not have a PDF, and without it I do not know how to calculate the moment generating function. ($X$ is also not discrete).



      I will appreciate some help.






      probability moment-generating-functions exponential-distribution







      share|cite|improve this question











      $endgroup$




      Let $Ysimtext{Exp}(1)$ be a random variable. I denote the random variable $X$ as $X=min(Y,1)$. The task is to find the moment generating function of $X$.



      By simply calculating the probability I managed to find the CDF of $X$ is:
      $$F_X(t) = begin{cases}
      1-e^{-t}, & 0 le t <1\
      1, & t ge 1\
      0, & text{otherwise}
      end{cases}$$



      Here I got stuck. Since $X$ is not a continuous variable, it does not have a PDF, and without it I do not know how to calculate the moment generating function. ($X$ is also not discrete).



      I will appreciate some help.






      probability moment-generating-functions exponential-distribution




      probability moment-generating-functions exponential-distribution






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 10 '18 at 9:47









      Saad

      19.7k92352




      19.7k92352










      asked Dec 10 '18 at 0:52









      Gabi GGabi G

      39819




      39819






















          1 Answer
          1






          active

          oldest

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          1












          $begingroup$

          The definition of the moment generating function does not require a PDF.
          $$phi_X(t) := E[e^{tX}].$$



          From here, it may be helpful to write
          $$e^{tX} = e^{tX} mathbf{1}{Y > 1} + e^{tX} mathbf{1}{Y le 1}$$
          and compute the expectation of the two terms separately.




          Edit for more details:



          $$E[e^{tX} mathbf{1}{Y > 1}] = E[e^{tY} mathbf{1}{Y > 1}] = int_1^infty e^{ty} f_Y(y) , dy = cdots.$$
          $$E[e^{tX} mathbf{1}{Y le 1}] = e^t E[mathbf{1}{Y le 1}] = e^t P(Y le 1) = cdots.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could you show me how?
            $endgroup$
            – Gabi G
            Dec 10 '18 at 7:25










          • $begingroup$
            Thank you! But I think the integral needs to be from 0 to 1
            $endgroup$
            – Gabi G
            Dec 10 '18 at 11:59










          • $begingroup$
            Thank you! But I think the integral needs to be from 0 to 1
            $endgroup$
            – Gabi G
            Dec 10 '18 at 11:59











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          1 Answer
          1






          active

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          active

          oldest

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          active

          oldest

          votes









          1












          $begingroup$

          The definition of the moment generating function does not require a PDF.
          $$phi_X(t) := E[e^{tX}].$$



          From here, it may be helpful to write
          $$e^{tX} = e^{tX} mathbf{1}{Y > 1} + e^{tX} mathbf{1}{Y le 1}$$
          and compute the expectation of the two terms separately.




          Edit for more details:



          $$E[e^{tX} mathbf{1}{Y > 1}] = E[e^{tY} mathbf{1}{Y > 1}] = int_1^infty e^{ty} f_Y(y) , dy = cdots.$$
          $$E[e^{tX} mathbf{1}{Y le 1}] = e^t E[mathbf{1}{Y le 1}] = e^t P(Y le 1) = cdots.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could you show me how?
            $endgroup$
            – Gabi G
            Dec 10 '18 at 7:25










          • $begingroup$
            Thank you! But I think the integral needs to be from 0 to 1
            $endgroup$
            – Gabi G
            Dec 10 '18 at 11:59










          • $begingroup$
            Thank you! But I think the integral needs to be from 0 to 1
            $endgroup$
            – Gabi G
            Dec 10 '18 at 11:59
















          1












          $begingroup$

          The definition of the moment generating function does not require a PDF.
          $$phi_X(t) := E[e^{tX}].$$



          From here, it may be helpful to write
          $$e^{tX} = e^{tX} mathbf{1}{Y > 1} + e^{tX} mathbf{1}{Y le 1}$$
          and compute the expectation of the two terms separately.




          Edit for more details:



          $$E[e^{tX} mathbf{1}{Y > 1}] = E[e^{tY} mathbf{1}{Y > 1}] = int_1^infty e^{ty} f_Y(y) , dy = cdots.$$
          $$E[e^{tX} mathbf{1}{Y le 1}] = e^t E[mathbf{1}{Y le 1}] = e^t P(Y le 1) = cdots.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could you show me how?
            $endgroup$
            – Gabi G
            Dec 10 '18 at 7:25










          • $begingroup$
            Thank you! But I think the integral needs to be from 0 to 1
            $endgroup$
            – Gabi G
            Dec 10 '18 at 11:59










          • $begingroup$
            Thank you! But I think the integral needs to be from 0 to 1
            $endgroup$
            – Gabi G
            Dec 10 '18 at 11:59














          1












          1








          1





          $begingroup$

          The definition of the moment generating function does not require a PDF.
          $$phi_X(t) := E[e^{tX}].$$



          From here, it may be helpful to write
          $$e^{tX} = e^{tX} mathbf{1}{Y > 1} + e^{tX} mathbf{1}{Y le 1}$$
          and compute the expectation of the two terms separately.




          Edit for more details:



          $$E[e^{tX} mathbf{1}{Y > 1}] = E[e^{tY} mathbf{1}{Y > 1}] = int_1^infty e^{ty} f_Y(y) , dy = cdots.$$
          $$E[e^{tX} mathbf{1}{Y le 1}] = e^t E[mathbf{1}{Y le 1}] = e^t P(Y le 1) = cdots.$$






          share|cite|improve this answer











          $endgroup$



          The definition of the moment generating function does not require a PDF.
          $$phi_X(t) := E[e^{tX}].$$



          From here, it may be helpful to write
          $$e^{tX} = e^{tX} mathbf{1}{Y > 1} + e^{tX} mathbf{1}{Y le 1}$$
          and compute the expectation of the two terms separately.




          Edit for more details:



          $$E[e^{tX} mathbf{1}{Y > 1}] = E[e^{tY} mathbf{1}{Y > 1}] = int_1^infty e^{ty} f_Y(y) , dy = cdots.$$
          $$E[e^{tX} mathbf{1}{Y le 1}] = e^t E[mathbf{1}{Y le 1}] = e^t P(Y le 1) = cdots.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 17:11

























          answered Dec 10 '18 at 1:00









          angryavianangryavian

          40.6k23380




          40.6k23380












          • $begingroup$
            Could you show me how?
            $endgroup$
            – Gabi G
            Dec 10 '18 at 7:25










          • $begingroup$
            Thank you! But I think the integral needs to be from 0 to 1
            $endgroup$
            – Gabi G
            Dec 10 '18 at 11:59










          • $begingroup$
            Thank you! But I think the integral needs to be from 0 to 1
            $endgroup$
            – Gabi G
            Dec 10 '18 at 11:59


















          • $begingroup$
            Could you show me how?
            $endgroup$
            – Gabi G
            Dec 10 '18 at 7:25










          • $begingroup$
            Thank you! But I think the integral needs to be from 0 to 1
            $endgroup$
            – Gabi G
            Dec 10 '18 at 11:59










          • $begingroup$
            Thank you! But I think the integral needs to be from 0 to 1
            $endgroup$
            – Gabi G
            Dec 10 '18 at 11:59
















          $begingroup$
          Could you show me how?
          $endgroup$
          – Gabi G
          Dec 10 '18 at 7:25




          $begingroup$
          Could you show me how?
          $endgroup$
          – Gabi G
          Dec 10 '18 at 7:25












          $begingroup$
          Thank you! But I think the integral needs to be from 0 to 1
          $endgroup$
          – Gabi G
          Dec 10 '18 at 11:59




          $begingroup$
          Thank you! But I think the integral needs to be from 0 to 1
          $endgroup$
          – Gabi G
          Dec 10 '18 at 11:59












          $begingroup$
          Thank you! But I think the integral needs to be from 0 to 1
          $endgroup$
          – Gabi G
          Dec 10 '18 at 11:59




          $begingroup$
          Thank you! But I think the integral needs to be from 0 to 1
          $endgroup$
          – Gabi G
          Dec 10 '18 at 11:59


















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