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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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
asked May 31 at 16:10
J. StottJ. Stott
285
$endgroup$
add a comment |
$begingroup$
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?
poisson-distribution
asked May 31 at 16:10
J. StottJ. Stott
285
$endgroup$
$begingroup$
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
add a comment |
$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?
poisson-distribution
asked May 31 at 16:10
J. StottJ. Stott
285
$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
poisson-distribution
asked May 31 at 16:10
J. StottJ. Stott
285
asked May 31 at 16:10
J. StottJ. Stott
285
asked May 31 at 16:10
J. StottJ. Stott
285
asked May 31 at 16:10
J. StottJ. Stott
285
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$
$$$$ "if a customer purchases 1 thing on average in a week, they purchase 4 things on average in a four-week period. Is this correct?" $textbfYes.$ $$$$ "Is this saying that the time between a customers purchases would grow exponentially as time goes on?" $textbfNo,$ that's not what it means at all. $qquad$
$endgroup$
– Michael Hardy
May 31 at 19:17
add a comment |
$begingroup$
For a full account of this, including definitions, see stats.stackexchange.com/questions/214421. For related posts, see stats.stackexchange.com/… Another useful search (focusing on "memoryless") is stats.stackexchange.com/….
$endgroup$
– whuber♦
May 31 at 16:29
$begingroup$
$$$$ "if a customer purchases 1 thing on average in a week, they purchase 4 things on average in a four-week period. Is this correct?" $textbfYes.$ $$$$ "Is this saying that the time between a customers purchases would grow exponentially as time goes on?" $textbfNo,$ that's not what it means at all. $qquad$
$endgroup$
– Michael Hardy
May 31 at 19:17
$begingroup$
For a full account of this, including definitions, see stats.stackexchange.com/questions/214421. For related posts, see stats.stackexchange.com/… Another useful search (focusing on "memoryless") is stats.stackexchange.com/….
$endgroup$
– whuber♦
May 31 at 16:29
$begingroup$
For a full account of this, including definitions, see stats.stackexchange.com/questions/214421. For related posts, see stats.stackexchange.com/… Another useful search (focusing on "memoryless") is stats.stackexchange.com/….
$endgroup$
– whuber♦
May 31 at 16:29
$begingroup$
$$$$ "if a customer purchases 1 thing on average in a week, they purchase 4 things on average in a four-week period. Is this correct?" $textbfYes.$ $$$$ "Is this saying that the time between a customers purchases would grow exponentially as time goes on?" $textbfNo,$ that's not what it means at all. $qquad$
$endgroup$
– Michael Hardy
May 31 at 19:17
$begingroup$
$$$$ "if a customer purchases 1 thing on average in a week, they purchase 4 things on average in a four-week period. Is this correct?" $textbfYes.$ $$$$ "Is this saying that the time between a customers purchases would grow exponentially as time goes on?" $textbfNo,$ that's not what it means at all. $qquad$
$endgroup$
– Michael Hardy
May 31 at 19:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $X_t$ be the number of arrivals in the Poisson process with rate $lambda$ between time $0$ and time $tge0.$ Then we have
$$
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.
$$
That says $T$ is exponentially distributed.
answered May 31 at 18:59
Michael HardyMichael Hardy
4,4541430
$endgroup$
$begingroup$
That's very clever. Thanks!
$endgroup$
– J. Stott
Jun 3 at 8:45
add a comment |
$begingroup$
Exponential distribution doesn't imply that time between events grows exponentially. All it tells you is that probability to wait longer between events declines very quickly with waiting time. The probability density is:$$lambda e^-lambda t$$ where $lambda$ is Poisson intensity, i.e. average number of events in unit of time, and $t$ is the waiting time. The average waiting time is obviously $1/lambda$.
answered May 31 at 16:15
AksakalAksakal
40.6k454123
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Let $X_t$ be the number of arrivals in the Poisson process with rate $lambda$ between time $0$ and time $tge0.$ Then we have
$$
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.
$$
That says $T$ is exponentially distributed.
answered May 31 at 18:59
Michael HardyMichael Hardy
4,4541430
$endgroup$
$begingroup$
That's very clever. Thanks!
$endgroup$
– J. Stott
Jun 3 at 8:45
add a comment |
$begingroup$
Let $X_t$ be the number of arrivals in the Poisson process with rate $lambda$ between time $0$ and time $tge0.$ Then we have
$$
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.
$$
That says $T$ is exponentially distributed.
answered May 31 at 18:59
Michael HardyMichael Hardy
4,4541430
$endgroup$
$begingroup$
That's very clever. Thanks!
$endgroup$
– J. Stott
Jun 3 at 8:45
add a comment |
$begingroup$
Let $X_t$ be the number of arrivals in the Poisson process with rate $lambda$ between time $0$ and time $tge0.$ Then we have
$$
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.
$$
That says $T$ is exponentially distributed.
answered May 31 at 18:59
Michael HardyMichael Hardy
4,4541430
$endgroup$
Let $X_t$ be the number of arrivals in the Poisson process with rate $lambda$ between time $0$ and time $tge0.$ Then we have
$$
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.
$$
That says $T$ is exponentially distributed.
answered May 31 at 18:59
Michael HardyMichael Hardy
4,4541430
answered May 31 at 18:59
Michael HardyMichael Hardy
4,4541430
answered May 31 at 18:59
Michael HardyMichael Hardy
4,4541430
answered May 31 at 18:59
Michael HardyMichael Hardy
4,4541430
4,4541430
$begingroup$
That's very clever. Thanks!
$endgroup$
– J. Stott
Jun 3 at 8:45
add a comment |
$begingroup$
That's very clever. Thanks!
$endgroup$
– J. Stott
Jun 3 at 8:45
$begingroup$
That's very clever. Thanks!
$endgroup$
– J. Stott
Jun 3 at 8:45
$begingroup$
That's very clever. Thanks!
$endgroup$
– J. Stott
Jun 3 at 8:45
add a comment |
$begingroup$
Exponential distribution doesn't imply that time between events grows exponentially. All it tells you is that probability to wait longer between events declines very quickly with waiting time. The probability density is:$$lambda e^-lambda t$$ where $lambda$ is Poisson intensity, i.e. average number of events in unit of time, and $t$ is the waiting time. The average waiting time is obviously $1/lambda$.
answered May 31 at 16:15
AksakalAksakal
40.6k454123
$endgroup$
add a comment |
$begingroup$
Exponential distribution doesn't imply that time between events grows exponentially. All it tells you is that probability to wait longer between events declines very quickly with waiting time. The probability density is:$$lambda e^-lambda t$$ where $lambda$ is Poisson intensity, i.e. average number of events in unit of time, and $t$ is the waiting time. The average waiting time is obviously $1/lambda$.
answered May 31 at 16:15
AksakalAksakal
40.6k454123
$endgroup$
add a comment |
$begingroup$
Exponential distribution doesn't imply that time between events grows exponentially. All it tells you is that probability to wait longer between events declines very quickly with waiting time. The probability density is:$$lambda e^-lambda t$$ where $lambda$ is Poisson intensity, i.e. average number of events in unit of time, and $t$ is the waiting time. The average waiting time is obviously $1/lambda$.
answered May 31 at 16:15
AksakalAksakal
40.6k454123
$endgroup$
Exponential distribution doesn't imply that time between events grows exponentially. All it tells you is that probability to wait longer between events declines very quickly with waiting time. The probability density is:$$lambda e^-lambda t$$ where $lambda$ is Poisson intensity, i.e. average number of events in unit of time, and $t$ is the waiting time. The average waiting time is obviously $1/lambda$.
answered May 31 at 16:15
AksakalAksakal
40.6k454123
answered May 31 at 16:15
AksakalAksakal
40.6k454123
answered May 31 at 16:15
AksakalAksakal
40.6k454123
answered May 31 at 16:15
AksakalAksakal
40.6k454123
40.6k454123
add a comment |
add a comment |
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$begingroup$
For a full account of this, including definitions, see stats.stackexchange.com/questions/214421. For related posts, see stats.stackexchange.com/… Another useful search (focusing on "memoryless") is stats.stackexchange.com/….
$endgroup$
– whuber♦
May 31 at 16:29
$begingroup$
$$$$ "if a customer purchases 1 thing on average in a week, they purchase 4 things on average in a four-week period. Is this correct?" $textbfYes.$ $$$$ "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