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Probability Distribution Problem

  1. Oct 4, 2011 #1
    1. The problem statement, all variables and given/known data

    Log-ons to a certain computer website occur randomly at a uniform average rate of 2.4 per minute. State the distribution of the number N of log-ons that occur during a period of t minutes.

    Obtain the probablity that at least one log on occurs during a period of t minutes.

    Hence obtain the probability density function of T, where T minutes is the interval between successive log-ons.

    Identify the distribution of T and state its mean and variance.

    2. Relevant equations



    3. The attempt at a solution

    This one's a bit embarrassing really. What I don't get is the "Hence obtain the probability density function of T, where T minutes is the interval between successive log-ons." part. It's only 3 marks for this section of the question so it's probably something simple. I must be missing some key fact or something.

    Here's what I did for the first bit.

    The distribution of N is Poisson with a mean of 2.4t so X~Po(2.4t)

    The probability of at least one log on is P(X≥1) = 1 - P(X=0) = 1 - e-2.4t

    And now I am stuck. This is from a really old A-Level Further Mathematics exam so I only had an examiner's report which gave the answers with a short and confusing explanation. It said that "The probability density function is obtained by taking the derivative of their answer 1 - e-2.4t. Which gives a negative exponential distribution 2.4e-2.4t. The mean and variance is easily 1/2.4 and (1/2.4)2 respectively."

    What I don't understand is why the probablity density function is the derivative of that expression.
     
  2. jcsd
  3. Oct 4, 2011 #2

    Ray Vickson

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    Let T1 = time to first log-on (after you start measuring times), T2 = time between first and second log-on, etc. The T1, T2, ... are the inter-arrival times of a Poisson process. Assuming the counting probabilities are independent between non-overlapping intervals, and that N = no. arrivals in (t1,t2) has the Poisson distribution with mean a*(t2-t1) [with a = your 2.4], we can show--by a lengthy but not particularly difficult argument--that T1, T2, ... are independent and identically distributed. Furthermore, the number arriving by time t is N(t), where Pr{N(t) = n} = Pr{T1+T2+...+Tn < t, T1+T2+...+Tn+T(n+1) > t} = (a*t)^n *exp(-a*t)/n!, where a = 2.4. In particular, Pr{T1 > t} = Pr{N(t)=0} = exp(-a*t), so the density f(t) of T1 satisfies f(t) = (d/dt) Pr{T1 > t} = a*exp(-a*t), because Pr{T1 > t} = integral_{x=t..infinity} f(x) dx. Also, T2, T3, T4,... all have the same density.

    RGV
     
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