Exponential Distribution Problem

AI Thread Summary
The discussion centers on solving an exponential distribution problem involving independent random variables X and Y, with the goal of determining if Z=min(X,Y) is also exponentially distributed and calculating its expectation. The probability that x < y can be derived using either the ratio X/Y or the difference X - Y, leveraging their independence. Additionally, the conversation shifts to representing an M/M/1 queue as a continuous-time Markov chain, where transition rates are defined by arrival and service rates. The participants explore how to characterize the states and transition probabilities within this queueing model, noting that the arrival rate is Poisson and the service rate is exponential. Overall, the discussion highlights the mathematical relationships and concepts necessary for understanding these statistical distributions and queueing theory.
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I am having trouble solving this problem. I'm not sure how to solve this problem... Assume X and Y are independent exponential random variables with means 1/x and 1/y, respectively. If Z=min(X,Y): Is Z exponentially distributed as well (if so, how do you know)? What is the expectation of Z? What is the probability that x < y?

Lastly, with the information from above, show how a M/M/1 queue could be represented as a Markov chain that is continuous-time with transition rates Qn,n+1=L and Qn,n-1=U, n=0,1,2,... M/M/1 queue=Arrival is Poisson/Service is Exponential/1 server (with an infinite buffer size).
 
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"Is Z exponentially distributed as well (if so, how do you know)?"

Prob {Z < z} = Prob {min(X,Y) < z} = 1 - Prob {min(X,Y) > z} = 1 - Prob {X > z and Y > z}. Using this logic, you can derive the distribution of Z and decide whether it looks exponential.

"What is the probability that x < y?"

You can use either route:

1. Prob {X < Y} = Prob {X/Y < 1}

2. Prob {X < Y} = Prob {X - Y < 0}

In either case, you need to derive the distribution of X/Y or X - Y.
 
Thanks, I'm working on that and it looks likes since X and Y are independent, I would be able to add them together in the denominator for the new mean (1/(x+y)).

Does anyone know about the Markov part? That sort of came out of nowhere.
 
A Markov chain describes a system that is in one state (out of two or more states) at each period (e.g., at the end of each day); the probability of going from state s today to state s' tomorrow is independent of yesterday's state. For MM1, what would those states be, and how would you characterize these transition probabilities?
 
For M/M/1, given that arrival rates is Poisson with mean L and service (exit) rate is Exponential with mean U, I would think that the probability of getting an additional person in the next state is L and losing a person in the next state is U, which is just from logic. I'm not sure how to compute that though...?
 
It appears as if you need to compute the probability of acquiring N new arrivals between the end of this period and the end of the next period, conditional on having acquired A arrivals by the end of this period. This will give you the transition probability between today's state (A) and tomorrow's (A+N).
 
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