Consider an arrival process whose interarrival times are independent exponential random variables with mean 2 (and consequently variance equal to 4), and consider the inter arrival interval IS seen by an observer who arrives at a fixed time t*, What is the variance of S ?
To find the variance of the inter-arrival interval seen by an observer who arrives at a fixed time t*, we need to use the properties of exponential distribution and the memorylessness property.
The inter-arrival times follow an exponential distribution with a mean of 2. This means that the rate parameter (λ) for the exponential distribution is equal to 1/2, which is reciprocal of the mean (1/mean).
Let's denote the random variable representing the inter-arrival interval seen by the observer as X. Since X is the sum of two independent exponential random variables, each with mean 2, we can express X as the sum of two exponential random variables with rate 1/2.
Using the memorylessness property of the exponential distribution, we know that the conditional distribution of X, given that the observer has already been waiting for time t*, is equivalent to the distribution of the sum of two independent exponential random variables with rate 1/2.
The sum of two independent exponential random variables with rate 1/2 follows a gamma distribution with parameters (2, 1/2). The variance of a gamma distribution with parameters (k, θ) is equal to kθ^2. Therefore, in this case, the variance of X is 2*(1/2)^2 = 1/2.
Hence, the variance of the inter-arrival interval seen by the observer who arrives at a fixed time t* is 1/2.