Probability Mass Function: Geometric Distribution over an interval












0












$begingroup$


Equation: Given a random variable $Z$, let $Z ~ Geometric(theta)$, Find $P(5 leq Z leq 9)$.



Attempt 1:



Try something like $P(5 leq Z leq 9) = P(Z = 9) - P(Z = 5) = theta[(1-theta)^9-(1-theta)^5]$



I know this is a valid method for continuous distributions, but I wasn't sure if it would work the same way for a discrete function, like geometric. So I tried to compute it manually in attempt #2.



Attempt 2:



$P(5 leq Z leq 9) = P(Z = 5) + P(Z = 6) + P(Z = 7) + P(Z = 8) + P(Z= 9) = $Some answer



I know this one works, but I feel it's a bit too much in terms of computation. Like, if I was given an equation that asked for an interval from $3$ to $1000$, then there's no way I could manually compute that by hand.



Which is why I was wondering if there was a more efficient method to calculate the geometric distribution over a given interval? I tried to put everything into a summation and derive an equation, but I get stuck after pulling the theta out of the equation such that it's



$sumlimits_{i=0}^n (1- theta)^itheta = $ $theta sumlimits_{i=0}^n (1- theta)^i $



Anybody have a better solution?






discrete-mathematics probability-distributions summation







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Equation: Given a random variable $Z$, let $Z ~ Geometric(theta)$, Find $P(5 leq Z leq 9)$.



    Attempt 1:



    Try something like $P(5 leq Z leq 9) = P(Z = 9) - P(Z = 5) = theta[(1-theta)^9-(1-theta)^5]$



    I know this is a valid method for continuous distributions, but I wasn't sure if it would work the same way for a discrete function, like geometric. So I tried to compute it manually in attempt #2.



    Attempt 2:



    $P(5 leq Z leq 9) = P(Z = 5) + P(Z = 6) + P(Z = 7) + P(Z = 8) + P(Z= 9) = $Some answer



    I know this one works, but I feel it's a bit too much in terms of computation. Like, if I was given an equation that asked for an interval from $3$ to $1000$, then there's no way I could manually compute that by hand.



    Which is why I was wondering if there was a more efficient method to calculate the geometric distribution over a given interval? I tried to put everything into a summation and derive an equation, but I get stuck after pulling the theta out of the equation such that it's



    $sumlimits_{i=0}^n (1- theta)^itheta = $ $theta sumlimits_{i=0}^n (1- theta)^i $



    Anybody have a better solution?






    discrete-mathematics probability-distributions summation







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Equation: Given a random variable $Z$, let $Z ~ Geometric(theta)$, Find $P(5 leq Z leq 9)$.



      Attempt 1:



      Try something like $P(5 leq Z leq 9) = P(Z = 9) - P(Z = 5) = theta[(1-theta)^9-(1-theta)^5]$



      I know this is a valid method for continuous distributions, but I wasn't sure if it would work the same way for a discrete function, like geometric. So I tried to compute it manually in attempt #2.



      Attempt 2:



      $P(5 leq Z leq 9) = P(Z = 5) + P(Z = 6) + P(Z = 7) + P(Z = 8) + P(Z= 9) = $Some answer



      I know this one works, but I feel it's a bit too much in terms of computation. Like, if I was given an equation that asked for an interval from $3$ to $1000$, then there's no way I could manually compute that by hand.



      Which is why I was wondering if there was a more efficient method to calculate the geometric distribution over a given interval? I tried to put everything into a summation and derive an equation, but I get stuck after pulling the theta out of the equation such that it's



      $sumlimits_{i=0}^n (1- theta)^itheta = $ $theta sumlimits_{i=0}^n (1- theta)^i $



      Anybody have a better solution?






      discrete-mathematics probability-distributions summation







      share|cite|improve this question











      $endgroup$




      Equation: Given a random variable $Z$, let $Z ~ Geometric(theta)$, Find $P(5 leq Z leq 9)$.



      Attempt 1:



      Try something like $P(5 leq Z leq 9) = P(Z = 9) - P(Z = 5) = theta[(1-theta)^9-(1-theta)^5]$



      I know this is a valid method for continuous distributions, but I wasn't sure if it would work the same way for a discrete function, like geometric. So I tried to compute it manually in attempt #2.



      Attempt 2:



      $P(5 leq Z leq 9) = P(Z = 5) + P(Z = 6) + P(Z = 7) + P(Z = 8) + P(Z= 9) = $Some answer



      I know this one works, but I feel it's a bit too much in terms of computation. Like, if I was given an equation that asked for an interval from $3$ to $1000$, then there's no way I could manually compute that by hand.



      Which is why I was wondering if there was a more efficient method to calculate the geometric distribution over a given interval? I tried to put everything into a summation and derive an equation, but I get stuck after pulling the theta out of the equation such that it's



      $sumlimits_{i=0}^n (1- theta)^itheta = $ $theta sumlimits_{i=0}^n (1- theta)^i $



      Anybody have a better solution?






      discrete-mathematics probability-distributions summation




      discrete-mathematics probability-distributions summation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 12 '18 at 22:59









      platty

      3,350320




      3,350320










      asked Dec 12 '18 at 22:08









      Donald MayerDonald Mayer

      336




      336






















          1 Answer
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          0












          $begingroup$

          Your original solution (first attempt) is not quite correct as written. Instead, you should be calculating $P(Z geq 5) - P(Z > 9)$. Intuitively, this is "the chance that $Z$ is at least $5$, but not more than $9$." For a geometric distribution, you could write this as $P(Z > 4) - P(Z > 9)$. But it is actually not too difficult to compute $P(Z > x)$ for any geometric distribution; this holds iff the first $x$ trials are failures, so it happens with probability $P(Z > x) = (1 - theta)^x$ (assuming $theta$ is the probability of success in a single trial). In general, this means that, for geometric $Z$, we have:
          $$
          begin{align*}
          P(a leq Z leq b) &= P(Z geq a) - P(Z > b) \
          &= P(Z > a-1) - P(Z > b) \
          &= (1-theta)^{a-1} - (1 - theta)^b
          end{align*}
          $$



          Also note that your summation (from the second attempt) works as well, once simplified. To find $P(a leq Z leq b)$, we note that, from the geometric distribution, this is exactly equal to
          $$sum_{k=a}^b theta (1 - theta)^{k-1}$$
          But this is a geometric series with initial term $theta(1 - theta)^{a-1}$ and common ratio $1 - theta$. The sum of such a finite geometric series with $b-a + 1$ terms is:



          $$
          begin{align*}
          frac{theta(1-theta)^{a-1} cdot left( 1 - (1 - theta)^{b-a + 1}right)}{1 - (1 - theta)} &= (1-theta)^{a-1} cdot left( 1 - (1 - theta)^{b-a + 1}right) \
          &= (1-theta)^{a-1} - (1 - theta)^b
          end{align*}
          $$
          So the two methods agree.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yes, thank you. I wasn't thinking straight. The first solution you provided is what i'm looking for.The second solution you provided is the same as the textbook answer, so thanks for explaining that as well.
            $endgroup$
            – Donald Mayer
            Dec 13 '18 at 15:06














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          $begingroup$

          Your original solution (first attempt) is not quite correct as written. Instead, you should be calculating $P(Z geq 5) - P(Z > 9)$. Intuitively, this is "the chance that $Z$ is at least $5$, but not more than $9$." For a geometric distribution, you could write this as $P(Z > 4) - P(Z > 9)$. But it is actually not too difficult to compute $P(Z > x)$ for any geometric distribution; this holds iff the first $x$ trials are failures, so it happens with probability $P(Z > x) = (1 - theta)^x$ (assuming $theta$ is the probability of success in a single trial). In general, this means that, for geometric $Z$, we have:
          $$
          begin{align*}
          P(a leq Z leq b) &= P(Z geq a) - P(Z > b) \
          &= P(Z > a-1) - P(Z > b) \
          &= (1-theta)^{a-1} - (1 - theta)^b
          end{align*}
          $$



          Also note that your summation (from the second attempt) works as well, once simplified. To find $P(a leq Z leq b)$, we note that, from the geometric distribution, this is exactly equal to
          $$sum_{k=a}^b theta (1 - theta)^{k-1}$$
          But this is a geometric series with initial term $theta(1 - theta)^{a-1}$ and common ratio $1 - theta$. The sum of such a finite geometric series with $b-a + 1$ terms is:



          $$
          begin{align*}
          frac{theta(1-theta)^{a-1} cdot left( 1 - (1 - theta)^{b-a + 1}right)}{1 - (1 - theta)} &= (1-theta)^{a-1} cdot left( 1 - (1 - theta)^{b-a + 1}right) \
          &= (1-theta)^{a-1} - (1 - theta)^b
          end{align*}
          $$
          So the two methods agree.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yes, thank you. I wasn't thinking straight. The first solution you provided is what i'm looking for.The second solution you provided is the same as the textbook answer, so thanks for explaining that as well.
            $endgroup$
            – Donald Mayer
            Dec 13 '18 at 15:06


















          0












          $begingroup$

          Your original solution (first attempt) is not quite correct as written. Instead, you should be calculating $P(Z geq 5) - P(Z > 9)$. Intuitively, this is "the chance that $Z$ is at least $5$, but not more than $9$." For a geometric distribution, you could write this as $P(Z > 4) - P(Z > 9)$. But it is actually not too difficult to compute $P(Z > x)$ for any geometric distribution; this holds iff the first $x$ trials are failures, so it happens with probability $P(Z > x) = (1 - theta)^x$ (assuming $theta$ is the probability of success in a single trial). In general, this means that, for geometric $Z$, we have:
          $$
          begin{align*}
          P(a leq Z leq b) &= P(Z geq a) - P(Z > b) \
          &= P(Z > a-1) - P(Z > b) \
          &= (1-theta)^{a-1} - (1 - theta)^b
          end{align*}
          $$



          Also note that your summation (from the second attempt) works as well, once simplified. To find $P(a leq Z leq b)$, we note that, from the geometric distribution, this is exactly equal to
          $$sum_{k=a}^b theta (1 - theta)^{k-1}$$
          But this is a geometric series with initial term $theta(1 - theta)^{a-1}$ and common ratio $1 - theta$. The sum of such a finite geometric series with $b-a + 1$ terms is:



          $$
          begin{align*}
          frac{theta(1-theta)^{a-1} cdot left( 1 - (1 - theta)^{b-a + 1}right)}{1 - (1 - theta)} &= (1-theta)^{a-1} cdot left( 1 - (1 - theta)^{b-a + 1}right) \
          &= (1-theta)^{a-1} - (1 - theta)^b
          end{align*}
          $$
          So the two methods agree.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yes, thank you. I wasn't thinking straight. The first solution you provided is what i'm looking for.The second solution you provided is the same as the textbook answer, so thanks for explaining that as well.
            $endgroup$
            – Donald Mayer
            Dec 13 '18 at 15:06
















          0












          0








          0





          $begingroup$

          Your original solution (first attempt) is not quite correct as written. Instead, you should be calculating $P(Z geq 5) - P(Z > 9)$. Intuitively, this is "the chance that $Z$ is at least $5$, but not more than $9$." For a geometric distribution, you could write this as $P(Z > 4) - P(Z > 9)$. But it is actually not too difficult to compute $P(Z > x)$ for any geometric distribution; this holds iff the first $x$ trials are failures, so it happens with probability $P(Z > x) = (1 - theta)^x$ (assuming $theta$ is the probability of success in a single trial). In general, this means that, for geometric $Z$, we have:
          $$
          begin{align*}
          P(a leq Z leq b) &= P(Z geq a) - P(Z > b) \
          &= P(Z > a-1) - P(Z > b) \
          &= (1-theta)^{a-1} - (1 - theta)^b
          end{align*}
          $$



          Also note that your summation (from the second attempt) works as well, once simplified. To find $P(a leq Z leq b)$, we note that, from the geometric distribution, this is exactly equal to
          $$sum_{k=a}^b theta (1 - theta)^{k-1}$$
          But this is a geometric series with initial term $theta(1 - theta)^{a-1}$ and common ratio $1 - theta$. The sum of such a finite geometric series with $b-a + 1$ terms is:



          $$
          begin{align*}
          frac{theta(1-theta)^{a-1} cdot left( 1 - (1 - theta)^{b-a + 1}right)}{1 - (1 - theta)} &= (1-theta)^{a-1} cdot left( 1 - (1 - theta)^{b-a + 1}right) \
          &= (1-theta)^{a-1} - (1 - theta)^b
          end{align*}
          $$
          So the two methods agree.






          share|cite|improve this answer











          $endgroup$



          Your original solution (first attempt) is not quite correct as written. Instead, you should be calculating $P(Z geq 5) - P(Z > 9)$. Intuitively, this is "the chance that $Z$ is at least $5$, but not more than $9$." For a geometric distribution, you could write this as $P(Z > 4) - P(Z > 9)$. But it is actually not too difficult to compute $P(Z > x)$ for any geometric distribution; this holds iff the first $x$ trials are failures, so it happens with probability $P(Z > x) = (1 - theta)^x$ (assuming $theta$ is the probability of success in a single trial). In general, this means that, for geometric $Z$, we have:
          $$
          begin{align*}
          P(a leq Z leq b) &= P(Z geq a) - P(Z > b) \
          &= P(Z > a-1) - P(Z > b) \
          &= (1-theta)^{a-1} - (1 - theta)^b
          end{align*}
          $$



          Also note that your summation (from the second attempt) works as well, once simplified. To find $P(a leq Z leq b)$, we note that, from the geometric distribution, this is exactly equal to
          $$sum_{k=a}^b theta (1 - theta)^{k-1}$$
          But this is a geometric series with initial term $theta(1 - theta)^{a-1}$ and common ratio $1 - theta$. The sum of such a finite geometric series with $b-a + 1$ terms is:



          $$
          begin{align*}
          frac{theta(1-theta)^{a-1} cdot left( 1 - (1 - theta)^{b-a + 1}right)}{1 - (1 - theta)} &= (1-theta)^{a-1} cdot left( 1 - (1 - theta)^{b-a + 1}right) \
          &= (1-theta)^{a-1} - (1 - theta)^b
          end{align*}
          $$
          So the two methods agree.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 12 '18 at 22:27

























          answered Dec 12 '18 at 22:15









          plattyplatty

          3,350320




          3,350320












          • $begingroup$
            Yes, thank you. I wasn't thinking straight. The first solution you provided is what i'm looking for.The second solution you provided is the same as the textbook answer, so thanks for explaining that as well.
            $endgroup$
            – Donald Mayer
            Dec 13 '18 at 15:06




















          • $begingroup$
            Yes, thank you. I wasn't thinking straight. The first solution you provided is what i'm looking for.The second solution you provided is the same as the textbook answer, so thanks for explaining that as well.
            $endgroup$
            – Donald Mayer
            Dec 13 '18 at 15:06


















          $begingroup$
          Yes, thank you. I wasn't thinking straight. The first solution you provided is what i'm looking for.The second solution you provided is the same as the textbook answer, so thanks for explaining that as well.
          $endgroup$
          – Donald Mayer
          Dec 13 '18 at 15:06






          $begingroup$
          Yes, thank you. I wasn't thinking straight. The first solution you provided is what i'm looking for.The second solution you provided is the same as the textbook answer, so thanks for explaining that as well.
          $endgroup$
          – Donald Mayer
          Dec 13 '18 at 15:06




















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