Poisson distribution: why does time between events follow an exponential distribution?The Number of Exponential Summands in a Fixed Interval is Poisson

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Poisson distribution: why does time between events follow an exponential distribution?


The Number of Exponential Summands in a Fixed Interval is Poisson






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3












$begingroup$


I was reading an article, and came across the following:




Purchase count follows a Poisson distribution with rate λ. In other
words, the timing of these purchases is somewhat random, but the rate
(in counts/unit time) is constant. In turn, this implies that the
inter-purchase time at the customer level should follow an exponential
distribution.




It's been quite a while since I did any statistics so I am struggling with the definitions of a Poisson distribution. What I understand by the "rate is constant" is that if a customer purchases 1 thing on average in a week, they purchase 4 things on average in a four-week period. Is this correct?



Where I believe I am confused is with the final sentence. Is this saying that the time between a customers purchases would grow exponentially as time goes on? Doesn't this contradict the idea that we have a constant rate of purchase?










share|cite|improve this question









$endgroup$











  • $begingroup$
    For a full account of this, including definitions, see stats.stackexchange.com/questions/214421. For related posts, see stats.stackexchange.com/… Another useful search (focusing on "memoryless") is stats.stackexchange.com/….
    $endgroup$
    – whuber
    May 31 at 16:29











  • $begingroup$
    $$$$ "if a customer purchases 1 thing on average in a week, they purchase 4 things on average in a four-week period. Is this correct?" $textbfYes.$ $$$$ "Is this saying that the time between a customers purchases would grow exponentially as time goes on?" $textbfNo,$ that's not what it means at all. $qquad$
    $endgroup$
    – Michael Hardy
    May 31 at 19:17

















3












$begingroup$


I was reading an article, and came across the following:




Purchase count follows a Poisson distribution with rate λ. In other
words, the timing of these purchases is somewhat random, but the rate
(in counts/unit time) is constant. In turn, this implies that the
inter-purchase time at the customer level should follow an exponential
distribution.




It's been quite a while since I did any statistics so I am struggling with the definitions of a Poisson distribution. What I understand by the "rate is constant" is that if a customer purchases 1 thing on average in a week, they purchase 4 things on average in a four-week period. Is this correct?



Where I believe I am confused is with the final sentence. Is this saying that the time between a customers purchases would grow exponentially as time goes on? Doesn't this contradict the idea that we have a constant rate of purchase?










share|cite|improve this question









$endgroup$











  • $begingroup$
    For a full account of this, including definitions, see stats.stackexchange.com/questions/214421. For related posts, see stats.stackexchange.com/… Another useful search (focusing on "memoryless") is stats.stackexchange.com/….
    $endgroup$
    – whuber
    May 31 at 16:29











  • $begingroup$
    $$$$ "if a customer purchases 1 thing on average in a week, they purchase 4 things on average in a four-week period. Is this correct?" $textbfYes.$ $$$$ "Is this saying that the time between a customers purchases would grow exponentially as time goes on?" $textbfNo,$ that's not what it means at all. $qquad$
    $endgroup$
    – Michael Hardy
    May 31 at 19:17













3












3








3


1



$begingroup$


I was reading an article, and came across the following:




Purchase count follows a Poisson distribution with rate λ. In other
words, the timing of these purchases is somewhat random, but the rate
(in counts/unit time) is constant. In turn, this implies that the
inter-purchase time at the customer level should follow an exponential
distribution.




It's been quite a while since I did any statistics so I am struggling with the definitions of a Poisson distribution. What I understand by the "rate is constant" is that if a customer purchases 1 thing on average in a week, they purchase 4 things on average in a four-week period. Is this correct?



Where I believe I am confused is with the final sentence. Is this saying that the time between a customers purchases would grow exponentially as time goes on? Doesn't this contradict the idea that we have a constant rate of purchase?










share|cite|improve this question









$endgroup$




I was reading an article, and came across the following:




Purchase count follows a Poisson distribution with rate λ. In other
words, the timing of these purchases is somewhat random, but the rate
(in counts/unit time) is constant. In turn, this implies that the
inter-purchase time at the customer level should follow an exponential
distribution.




It's been quite a while since I did any statistics so I am struggling with the definitions of a Poisson distribution. What I understand by the "rate is constant" is that if a customer purchases 1 thing on average in a week, they purchase 4 things on average in a four-week period. Is this correct?



Where I believe I am confused is with the final sentence. Is this saying that the time between a customers purchases would grow exponentially as time goes on? Doesn't this contradict the idea that we have a constant rate of purchase?







poisson-distribution






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asked May 31 at 16:10









J. StottJ. Stott

285




285











  • $begingroup$
    For a full account of this, including definitions, see stats.stackexchange.com/questions/214421. For related posts, see stats.stackexchange.com/… Another useful search (focusing on "memoryless") is stats.stackexchange.com/….
    $endgroup$
    – whuber
    May 31 at 16:29











  • $begingroup$
    $$$$ "if a customer purchases 1 thing on average in a week, they purchase 4 things on average in a four-week period. Is this correct?" $textbfYes.$ $$$$ "Is this saying that the time between a customers purchases would grow exponentially as time goes on?" $textbfNo,$ that's not what it means at all. $qquad$
    $endgroup$
    – Michael Hardy
    May 31 at 19:17
















  • $begingroup$
    For a full account of this, including definitions, see stats.stackexchange.com/questions/214421. For related posts, see stats.stackexchange.com/… Another useful search (focusing on "memoryless") is stats.stackexchange.com/….
    $endgroup$
    – whuber
    May 31 at 16:29











  • $begingroup$
    $$$$ "if a customer purchases 1 thing on average in a week, they purchase 4 things on average in a four-week period. Is this correct?" $textbfYes.$ $$$$ "Is this saying that the time between a customers purchases would grow exponentially as time goes on?" $textbfNo,$ that's not what it means at all. $qquad$
    $endgroup$
    – Michael Hardy
    May 31 at 19:17















$begingroup$
For a full account of this, including definitions, see stats.stackexchange.com/questions/214421. For related posts, see stats.stackexchange.com/… Another useful search (focusing on "memoryless") is stats.stackexchange.com/….
$endgroup$
– whuber
May 31 at 16:29





$begingroup$
For a full account of this, including definitions, see stats.stackexchange.com/questions/214421. For related posts, see stats.stackexchange.com/… Another useful search (focusing on "memoryless") is stats.stackexchange.com/….
$endgroup$
– whuber
May 31 at 16:29













$begingroup$
$$$$ "if a customer purchases 1 thing on average in a week, they purchase 4 things on average in a four-week period. Is this correct?" $textbfYes.$ $$$$ "Is this saying that the time between a customers purchases would grow exponentially as time goes on?" $textbfNo,$ that's not what it means at all. $qquad$
$endgroup$
– Michael Hardy
May 31 at 19:17




$begingroup$
$$$$ "if a customer purchases 1 thing on average in a week, they purchase 4 things on average in a four-week period. Is this correct?" $textbfYes.$ $$$$ "Is this saying that the time between a customers purchases would grow exponentially as time goes on?" $textbfNo,$ that's not what it means at all. $qquad$
$endgroup$
– Michael Hardy
May 31 at 19:17










2 Answers
2






active

oldest

votes


















3












$begingroup$

Let $X_t$ be the number of arrivals in the Poisson process with rate $lambda$ between time $0$ and time $tge0.$ Then we have
$$
Pr(X_t=x) = frac(lambda t)^x e^-lambda tx! text for x=0,1,2,3,ldots
$$



Let $T$ be the time until the first arrival.



Then the following two events are really both the same event:
$$
Big[ X_t=0Big]. qquad Big[ T>t Big].
$$

So they both have the same probability. Thus
$$
Pr(T>t) = Pr(X_t=0) = frac(lambda t)^0 e^-lambda t0! = e^-lambda t.
$$

So
$$
Pr(T>t) = e^-lambda t text for tge0.
$$

That says $T$ is exponentially distributed.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    That's very clever. Thanks!
    $endgroup$
    – J. Stott
    Jun 3 at 8:45


















3












$begingroup$

Exponential distribution doesn't imply that time between events grows exponentially. All it tells you is that probability to wait longer between events declines very quickly with waiting time. The probability density is:$$lambda e^-lambda t$$ where $lambda$ is Poisson intensity, i.e. average number of events in unit of time, and $t$ is the waiting time. The average waiting time is obviously $1/lambda$.






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    2 Answers
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    active

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    2 Answers
    2






    active

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    active

    oldest

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    3












    $begingroup$

    Let $X_t$ be the number of arrivals in the Poisson process with rate $lambda$ between time $0$ and time $tge0.$ Then we have
    $$
    Pr(X_t=x) = frac(lambda t)^x e^-lambda tx! text for x=0,1,2,3,ldots
    $$



    Let $T$ be the time until the first arrival.



    Then the following two events are really both the same event:
    $$
    Big[ X_t=0Big]. qquad Big[ T>t Big].
    $$

    So they both have the same probability. Thus
    $$
    Pr(T>t) = Pr(X_t=0) = frac(lambda t)^0 e^-lambda t0! = e^-lambda t.
    $$

    So
    $$
    Pr(T>t) = e^-lambda t text for tge0.
    $$

    That says $T$ is exponentially distributed.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      That's very clever. Thanks!
      $endgroup$
      – J. Stott
      Jun 3 at 8:45















    3












    $begingroup$

    Let $X_t$ be the number of arrivals in the Poisson process with rate $lambda$ between time $0$ and time $tge0.$ Then we have
    $$
    Pr(X_t=x) = frac(lambda t)^x e^-lambda tx! text for x=0,1,2,3,ldots
    $$



    Let $T$ be the time until the first arrival.



    Then the following two events are really both the same event:
    $$
    Big[ X_t=0Big]. qquad Big[ T>t Big].
    $$

    So they both have the same probability. Thus
    $$
    Pr(T>t) = Pr(X_t=0) = frac(lambda t)^0 e^-lambda t0! = e^-lambda t.
    $$

    So
    $$
    Pr(T>t) = e^-lambda t text for tge0.
    $$

    That says $T$ is exponentially distributed.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      That's very clever. Thanks!
      $endgroup$
      – J. Stott
      Jun 3 at 8:45













    3












    3








    3





    $begingroup$

    Let $X_t$ be the number of arrivals in the Poisson process with rate $lambda$ between time $0$ and time $tge0.$ Then we have
    $$
    Pr(X_t=x) = frac(lambda t)^x e^-lambda tx! text for x=0,1,2,3,ldots
    $$



    Let $T$ be the time until the first arrival.



    Then the following two events are really both the same event:
    $$
    Big[ X_t=0Big]. qquad Big[ T>t Big].
    $$

    So they both have the same probability. Thus
    $$
    Pr(T>t) = Pr(X_t=0) = frac(lambda t)^0 e^-lambda t0! = e^-lambda t.
    $$

    So
    $$
    Pr(T>t) = e^-lambda t text for tge0.
    $$

    That says $T$ is exponentially distributed.






    share|cite|improve this answer









    $endgroup$



    Let $X_t$ be the number of arrivals in the Poisson process with rate $lambda$ between time $0$ and time $tge0.$ Then we have
    $$
    Pr(X_t=x) = frac(lambda t)^x e^-lambda tx! text for x=0,1,2,3,ldots
    $$



    Let $T$ be the time until the first arrival.



    Then the following two events are really both the same event:
    $$
    Big[ X_t=0Big]. qquad Big[ T>t Big].
    $$

    So they both have the same probability. Thus
    $$
    Pr(T>t) = Pr(X_t=0) = frac(lambda t)^0 e^-lambda t0! = e^-lambda t.
    $$

    So
    $$
    Pr(T>t) = e^-lambda t text for tge0.
    $$

    That says $T$ is exponentially distributed.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered May 31 at 18:59









    Michael HardyMichael Hardy

    4,4541430




    4,4541430











    • $begingroup$
      That's very clever. Thanks!
      $endgroup$
      – J. Stott
      Jun 3 at 8:45
















    • $begingroup$
      That's very clever. Thanks!
      $endgroup$
      – J. Stott
      Jun 3 at 8:45















    $begingroup$
    That's very clever. Thanks!
    $endgroup$
    – J. Stott
    Jun 3 at 8:45




    $begingroup$
    That's very clever. Thanks!
    $endgroup$
    – J. Stott
    Jun 3 at 8:45













    3












    $begingroup$

    Exponential distribution doesn't imply that time between events grows exponentially. All it tells you is that probability to wait longer between events declines very quickly with waiting time. The probability density is:$$lambda e^-lambda t$$ where $lambda$ is Poisson intensity, i.e. average number of events in unit of time, and $t$ is the waiting time. The average waiting time is obviously $1/lambda$.






    share|cite|improve this answer









    $endgroup$

















      3












      $begingroup$

      Exponential distribution doesn't imply that time between events grows exponentially. All it tells you is that probability to wait longer between events declines very quickly with waiting time. The probability density is:$$lambda e^-lambda t$$ where $lambda$ is Poisson intensity, i.e. average number of events in unit of time, and $t$ is the waiting time. The average waiting time is obviously $1/lambda$.






      share|cite|improve this answer









      $endgroup$















        3












        3








        3





        $begingroup$

        Exponential distribution doesn't imply that time between events grows exponentially. All it tells you is that probability to wait longer between events declines very quickly with waiting time. The probability density is:$$lambda e^-lambda t$$ where $lambda$ is Poisson intensity, i.e. average number of events in unit of time, and $t$ is the waiting time. The average waiting time is obviously $1/lambda$.






        share|cite|improve this answer









        $endgroup$



        Exponential distribution doesn't imply that time between events grows exponentially. All it tells you is that probability to wait longer between events declines very quickly with waiting time. The probability density is:$$lambda e^-lambda t$$ where $lambda$ is Poisson intensity, i.e. average number of events in unit of time, and $t$ is the waiting time. The average waiting time is obviously $1/lambda$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered May 31 at 16:15









        AksakalAksakal

        40.6k454123




        40.6k454123



























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