HallsofIvy said:
The original statement is badly phrased. It is NOT correct that " the probability of each trial would approach zero." The probability of "each trial" is zero.
Well: strictly speaking, the statement "the probability of each trial" is nonsense - you can have a probability of a particular "outcome" of each trial. Each trial may have many outcomes, each with their own probability.
However, in a continuous situation like this you would never ask about individual trials, you would only ask about intervals.
I concur.
In a continuous distribution, the probability of an outcome would depend on an interval of the probability density function for that outcome.
eterna said:
From a site introducing the Poisson Distribution
... we really need a link to the site in question - it could just be that the site is rubbish.
[edit]... looks like it may be this one:
http://www.milefoot.com/math/stat/pdfd-poisson.htm [edit]
The question revisited:
...can someone explain why the probability of success for each trial would approach zero? Isn't it fixed?
... short answer: no - it is not fixed.
But context is everything.
[edit] using the above website: the quoted passage does not refer to the poisson distribution - the second half of the passage spells out that the rate of successes is a constant, but the probability of a particular number of successes in an interval is not.
The passage being considered is:
"When trials can occur in a fixed continuum of time (or distance), each instant of time (or distance) is essentially a distinct trial. Because a continuum contains an infinity of points, this means a statistical experiment may have an infinite number of trials, and the probability of each trial would approach zero."
I don't think the passage quoted is actually supposed to refer to any special property of the Poisson distribution.
I think the author is trying to make a connection between discrete and continuous probabilities.
Consider:
Say I roll a die every minute:
If N is the number of sixes in an hour, then ##P(N=n)\sim \text{Pois}(n;10)##: $$p(n)=\frac{10^ne^{-10}}{n!}$$
A single trial would, therefore, be a single hour-long observation.
The author seems to be saying that if the trial interval is shorter, then p(n) gets smaller.
If I make the trial interval T shorter, then $$p(n;T)=\frac{(T/6)^ne^{-T/6}}{n!}$$... if T is in minutes.
... the effect is to make small values of n more likely as T decreases: how many sixes would I expect in 10 seconds, 5, 1?
Imagine we define a "success" as the event that N=0 (i.e. no sixes are rolled in the time interval)
Then the probability of a success, ##p(0;T)=e^{-T/6}##, approaches 1 and T approaches 0.
... this seems at odds with the passage quoted - which is why I suspect this passage must come before the Poisson distribution is actually introduced. Either that or the author got confused.
Certainly the statement requires a lot of qualification to make it work sensibly.
So how about it eterna: any of this help?
Still confused: please supply link.
[edit]... if the link is as above, then the whole passage goes like this:
When trials can occur in a fixed continuum of time (or distance), each instant of time (or distance) is essentially a distinct trial. Because a continuum contains an infinity of points, this means a statistical experiment may have an infinite number of trials, and the probability of each trial would approach zero. But the rate of successes per unit of time (or distance) can still be a finite, nonzero quantity.
Let us assume that there are only two possible outcomes for each instant of time (or distance), that two successes cannot occur at the same instant, that trials are independent, and that the rate of successes is constant. Let X represent the number of successes. Then X has a Poisson distribution.
... you left out the bit not in italics.
I also separated out the passage into two paragraphs to help clarity.
The first paragraph is more general concerning continuous distributions and the second attempts to motivate the Poisson distribution.
