Finding constant from probability mass function












0












$begingroup$


Let $X$ be a discrete random variable with probability distribution (probability mass function),



$P(x) = c(frac{1}{3})^x, x = 0,1,2,...$




a) Find $c$ such that $P(x)$ is a legitimate PDF.



b) Find the CDF of $X$, $F(x), x ∈ {0,1,2,...}$




b) Here I was thinking of using this formula:



$sum_{k = 1}^x P(k)$



So,



$sum_{k = 0}^x c(frac{1}{3})^k = sum_{k = 1}^x c(frac{1}{3})^{x - 1}$



$= cfrac{(1 - frac{1}{3}^x)}{1 - frac{1}{3}}$



Does it even make sense?



I'm stuck on a)






probability functions random-variables







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let $X$ be a discrete random variable with probability distribution (probability mass function),



    $P(x) = c(frac{1}{3})^x, x = 0,1,2,...$




    a) Find $c$ such that $P(x)$ is a legitimate PDF.



    b) Find the CDF of $X$, $F(x), x ∈ {0,1,2,...}$




    b) Here I was thinking of using this formula:



    $sum_{k = 1}^x P(k)$



    So,



    $sum_{k = 0}^x c(frac{1}{3})^k = sum_{k = 1}^x c(frac{1}{3})^{x - 1}$



    $= cfrac{(1 - frac{1}{3}^x)}{1 - frac{1}{3}}$



    Does it even make sense?



    I'm stuck on a)






    probability functions random-variables







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $X$ be a discrete random variable with probability distribution (probability mass function),



      $P(x) = c(frac{1}{3})^x, x = 0,1,2,...$




      a) Find $c$ such that $P(x)$ is a legitimate PDF.



      b) Find the CDF of $X$, $F(x), x ∈ {0,1,2,...}$




      b) Here I was thinking of using this formula:



      $sum_{k = 1}^x P(k)$



      So,



      $sum_{k = 0}^x c(frac{1}{3})^k = sum_{k = 1}^x c(frac{1}{3})^{x - 1}$



      $= cfrac{(1 - frac{1}{3}^x)}{1 - frac{1}{3}}$



      Does it even make sense?



      I'm stuck on a)






      probability functions random-variables







      share|cite|improve this question









      $endgroup$




      Let $X$ be a discrete random variable with probability distribution (probability mass function),



      $P(x) = c(frac{1}{3})^x, x = 0,1,2,...$




      a) Find $c$ such that $P(x)$ is a legitimate PDF.



      b) Find the CDF of $X$, $F(x), x ∈ {0,1,2,...}$




      b) Here I was thinking of using this formula:



      $sum_{k = 1}^x P(k)$



      So,



      $sum_{k = 0}^x c(frac{1}{3})^k = sum_{k = 1}^x c(frac{1}{3})^{x - 1}$



      $= cfrac{(1 - frac{1}{3}^x)}{1 - frac{1}{3}}$



      Does it even make sense?



      I'm stuck on a)






      probability functions random-variables




      probability functions random-variables






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Oct 3 '17 at 19:34









      HelloHello

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          $begingroup$

          To find $c$ we must have
          $$
          sum_{x=0}^{infty}P(x)=1quadLongrightarrow quadsum_{x=0}^{infty}cleft(frac 13right)^x=cfrac32=1quadLongrightarrow quad c=frac23
          $$



          For b) we have



          $$
          F(x)=sum_{k=0}^{x}P(x)=frac23sum_{k=0}^{x}left(frac 13right)^k=1-frac{1}{3^{x+1}}qquad text{for};x=0,1,2,ldots
          $$
          and $0$ elsewhere.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How do you go from $sum_{x=0}^{infty}cleft(frac 13right)^x$ to $cfrac32$?
            $endgroup$
            – Hello
            Oct 3 '17 at 19:44






          • 1




            $begingroup$
            This is a geometric series: $sum_{k=0}^infty r^k=frac{1}{1-r}$ for $|r|<1$
            $endgroup$
            – alexjo
            Oct 3 '17 at 19:45












          • $begingroup$
            I know that. I'm just a little confused as to how to plug in infinity
            $endgroup$
            – Hello
            Oct 3 '17 at 19:48










          • $begingroup$
            For $r=frac 13$ you have $sum_{k=0}^{infty}c(1/3)^k=csum_{k=0}^{infty}(1/3)^k=frac{c}{1-frac 13}=frac{c}{frac 23}=1$ and then $c=frac 23$.
            $endgroup$
            – alexjo
            Oct 3 '17 at 19:51












          • $begingroup$
            The partial sum is $sum_{k=0}^{n}r^k=frac{1-r^{n+1}}{1-r}$ and taking the limit for $ntoinfty$ with $|r|<1$, $r^{n+1}to 0$ and then $$sum_{k=0}^{infty}r^k=lim_{ntoinfty}sum_{k=0}^{n}r^k=lim_{ntoinfty}frac{1-r^{n+1}}{1-r}=frac{1}{1-r}qquad |r|<1$$
            $endgroup$
            – alexjo
            Oct 3 '17 at 19:56













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          1 Answer
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          0












          $begingroup$

          To find $c$ we must have
          $$
          sum_{x=0}^{infty}P(x)=1quadLongrightarrow quadsum_{x=0}^{infty}cleft(frac 13right)^x=cfrac32=1quadLongrightarrow quad c=frac23
          $$



          For b) we have



          $$
          F(x)=sum_{k=0}^{x}P(x)=frac23sum_{k=0}^{x}left(frac 13right)^k=1-frac{1}{3^{x+1}}qquad text{for};x=0,1,2,ldots
          $$
          and $0$ elsewhere.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How do you go from $sum_{x=0}^{infty}cleft(frac 13right)^x$ to $cfrac32$?
            $endgroup$
            – Hello
            Oct 3 '17 at 19:44






          • 1




            $begingroup$
            This is a geometric series: $sum_{k=0}^infty r^k=frac{1}{1-r}$ for $|r|<1$
            $endgroup$
            – alexjo
            Oct 3 '17 at 19:45












          • $begingroup$
            I know that. I'm just a little confused as to how to plug in infinity
            $endgroup$
            – Hello
            Oct 3 '17 at 19:48










          • $begingroup$
            For $r=frac 13$ you have $sum_{k=0}^{infty}c(1/3)^k=csum_{k=0}^{infty}(1/3)^k=frac{c}{1-frac 13}=frac{c}{frac 23}=1$ and then $c=frac 23$.
            $endgroup$
            – alexjo
            Oct 3 '17 at 19:51












          • $begingroup$
            The partial sum is $sum_{k=0}^{n}r^k=frac{1-r^{n+1}}{1-r}$ and taking the limit for $ntoinfty$ with $|r|<1$, $r^{n+1}to 0$ and then $$sum_{k=0}^{infty}r^k=lim_{ntoinfty}sum_{k=0}^{n}r^k=lim_{ntoinfty}frac{1-r^{n+1}}{1-r}=frac{1}{1-r}qquad |r|<1$$
            $endgroup$
            – alexjo
            Oct 3 '17 at 19:56


















          0












          $begingroup$

          To find $c$ we must have
          $$
          sum_{x=0}^{infty}P(x)=1quadLongrightarrow quadsum_{x=0}^{infty}cleft(frac 13right)^x=cfrac32=1quadLongrightarrow quad c=frac23
          $$



          For b) we have



          $$
          F(x)=sum_{k=0}^{x}P(x)=frac23sum_{k=0}^{x}left(frac 13right)^k=1-frac{1}{3^{x+1}}qquad text{for};x=0,1,2,ldots
          $$
          and $0$ elsewhere.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How do you go from $sum_{x=0}^{infty}cleft(frac 13right)^x$ to $cfrac32$?
            $endgroup$
            – Hello
            Oct 3 '17 at 19:44






          • 1




            $begingroup$
            This is a geometric series: $sum_{k=0}^infty r^k=frac{1}{1-r}$ for $|r|<1$
            $endgroup$
            – alexjo
            Oct 3 '17 at 19:45












          • $begingroup$
            I know that. I'm just a little confused as to how to plug in infinity
            $endgroup$
            – Hello
            Oct 3 '17 at 19:48










          • $begingroup$
            For $r=frac 13$ you have $sum_{k=0}^{infty}c(1/3)^k=csum_{k=0}^{infty}(1/3)^k=frac{c}{1-frac 13}=frac{c}{frac 23}=1$ and then $c=frac 23$.
            $endgroup$
            – alexjo
            Oct 3 '17 at 19:51












          • $begingroup$
            The partial sum is $sum_{k=0}^{n}r^k=frac{1-r^{n+1}}{1-r}$ and taking the limit for $ntoinfty$ with $|r|<1$, $r^{n+1}to 0$ and then $$sum_{k=0}^{infty}r^k=lim_{ntoinfty}sum_{k=0}^{n}r^k=lim_{ntoinfty}frac{1-r^{n+1}}{1-r}=frac{1}{1-r}qquad |r|<1$$
            $endgroup$
            – alexjo
            Oct 3 '17 at 19:56
















          0












          0








          0





          $begingroup$

          To find $c$ we must have
          $$
          sum_{x=0}^{infty}P(x)=1quadLongrightarrow quadsum_{x=0}^{infty}cleft(frac 13right)^x=cfrac32=1quadLongrightarrow quad c=frac23
          $$



          For b) we have



          $$
          F(x)=sum_{k=0}^{x}P(x)=frac23sum_{k=0}^{x}left(frac 13right)^k=1-frac{1}{3^{x+1}}qquad text{for};x=0,1,2,ldots
          $$
          and $0$ elsewhere.






          share|cite|improve this answer











          $endgroup$



          To find $c$ we must have
          $$
          sum_{x=0}^{infty}P(x)=1quadLongrightarrow quadsum_{x=0}^{infty}cleft(frac 13right)^x=cfrac32=1quadLongrightarrow quad c=frac23
          $$



          For b) we have



          $$
          F(x)=sum_{k=0}^{x}P(x)=frac23sum_{k=0}^{x}left(frac 13right)^k=1-frac{1}{3^{x+1}}qquad text{for};x=0,1,2,ldots
          $$
          and $0$ elsewhere.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Oct 3 '17 at 20:08

























          answered Oct 3 '17 at 19:38









          alexjoalexjo

          12.4k1430




          12.4k1430












          • $begingroup$
            How do you go from $sum_{x=0}^{infty}cleft(frac 13right)^x$ to $cfrac32$?
            $endgroup$
            – Hello
            Oct 3 '17 at 19:44






          • 1




            $begingroup$
            This is a geometric series: $sum_{k=0}^infty r^k=frac{1}{1-r}$ for $|r|<1$
            $endgroup$
            – alexjo
            Oct 3 '17 at 19:45












          • $begingroup$
            I know that. I'm just a little confused as to how to plug in infinity
            $endgroup$
            – Hello
            Oct 3 '17 at 19:48










          • $begingroup$
            For $r=frac 13$ you have $sum_{k=0}^{infty}c(1/3)^k=csum_{k=0}^{infty}(1/3)^k=frac{c}{1-frac 13}=frac{c}{frac 23}=1$ and then $c=frac 23$.
            $endgroup$
            – alexjo
            Oct 3 '17 at 19:51












          • $begingroup$
            The partial sum is $sum_{k=0}^{n}r^k=frac{1-r^{n+1}}{1-r}$ and taking the limit for $ntoinfty$ with $|r|<1$, $r^{n+1}to 0$ and then $$sum_{k=0}^{infty}r^k=lim_{ntoinfty}sum_{k=0}^{n}r^k=lim_{ntoinfty}frac{1-r^{n+1}}{1-r}=frac{1}{1-r}qquad |r|<1$$
            $endgroup$
            – alexjo
            Oct 3 '17 at 19:56




















          • $begingroup$
            How do you go from $sum_{x=0}^{infty}cleft(frac 13right)^x$ to $cfrac32$?
            $endgroup$
            – Hello
            Oct 3 '17 at 19:44






          • 1




            $begingroup$
            This is a geometric series: $sum_{k=0}^infty r^k=frac{1}{1-r}$ for $|r|<1$
            $endgroup$
            – alexjo
            Oct 3 '17 at 19:45












          • $begingroup$
            I know that. I'm just a little confused as to how to plug in infinity
            $endgroup$
            – Hello
            Oct 3 '17 at 19:48










          • $begingroup$
            For $r=frac 13$ you have $sum_{k=0}^{infty}c(1/3)^k=csum_{k=0}^{infty}(1/3)^k=frac{c}{1-frac 13}=frac{c}{frac 23}=1$ and then $c=frac 23$.
            $endgroup$
            – alexjo
            Oct 3 '17 at 19:51












          • $begingroup$
            The partial sum is $sum_{k=0}^{n}r^k=frac{1-r^{n+1}}{1-r}$ and taking the limit for $ntoinfty$ with $|r|<1$, $r^{n+1}to 0$ and then $$sum_{k=0}^{infty}r^k=lim_{ntoinfty}sum_{k=0}^{n}r^k=lim_{ntoinfty}frac{1-r^{n+1}}{1-r}=frac{1}{1-r}qquad |r|<1$$
            $endgroup$
            – alexjo
            Oct 3 '17 at 19:56


















          $begingroup$
          How do you go from $sum_{x=0}^{infty}cleft(frac 13right)^x$ to $cfrac32$?
          $endgroup$
          – Hello
          Oct 3 '17 at 19:44




          $begingroup$
          How do you go from $sum_{x=0}^{infty}cleft(frac 13right)^x$ to $cfrac32$?
          $endgroup$
          – Hello
          Oct 3 '17 at 19:44




          1




          1




          $begingroup$
          This is a geometric series: $sum_{k=0}^infty r^k=frac{1}{1-r}$ for $|r|<1$
          $endgroup$
          – alexjo
          Oct 3 '17 at 19:45






          $begingroup$
          This is a geometric series: $sum_{k=0}^infty r^k=frac{1}{1-r}$ for $|r|<1$
          $endgroup$
          – alexjo
          Oct 3 '17 at 19:45














          $begingroup$
          I know that. I'm just a little confused as to how to plug in infinity
          $endgroup$
          – Hello
          Oct 3 '17 at 19:48




          $begingroup$
          I know that. I'm just a little confused as to how to plug in infinity
          $endgroup$
          – Hello
          Oct 3 '17 at 19:48












          $begingroup$
          For $r=frac 13$ you have $sum_{k=0}^{infty}c(1/3)^k=csum_{k=0}^{infty}(1/3)^k=frac{c}{1-frac 13}=frac{c}{frac 23}=1$ and then $c=frac 23$.
          $endgroup$
          – alexjo
          Oct 3 '17 at 19:51






          $begingroup$
          For $r=frac 13$ you have $sum_{k=0}^{infty}c(1/3)^k=csum_{k=0}^{infty}(1/3)^k=frac{c}{1-frac 13}=frac{c}{frac 23}=1$ and then $c=frac 23$.
          $endgroup$
          – alexjo
          Oct 3 '17 at 19:51














          $begingroup$
          The partial sum is $sum_{k=0}^{n}r^k=frac{1-r^{n+1}}{1-r}$ and taking the limit for $ntoinfty$ with $|r|<1$, $r^{n+1}to 0$ and then $$sum_{k=0}^{infty}r^k=lim_{ntoinfty}sum_{k=0}^{n}r^k=lim_{ntoinfty}frac{1-r^{n+1}}{1-r}=frac{1}{1-r}qquad |r|<1$$
          $endgroup$
          – alexjo
          Oct 3 '17 at 19:56






          $begingroup$
          The partial sum is $sum_{k=0}^{n}r^k=frac{1-r^{n+1}}{1-r}$ and taking the limit for $ntoinfty$ with $|r|<1$, $r^{n+1}to 0$ and then $$sum_{k=0}^{infty}r^k=lim_{ntoinfty}sum_{k=0}^{n}r^k=lim_{ntoinfty}frac{1-r^{n+1}}{1-r}=frac{1}{1-r}qquad |r|<1$$
          $endgroup$
          – alexjo
          Oct 3 '17 at 19:56




















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