Exponential Distribution Problem

In summary, the conversation discusses the distribution of Z, the probability of X being less than Y, and how a M/M/1 queue can be represented as a Markov chain. It is mentioned that Z is exponentially distributed due to the independence of X and Y, and the probability of X being less than Y can be calculated using the distribution of X/Y or X-Y. The Markov chain for M/M/1 is described as a system where the arrival and service rates determine the probabilities of transitioning between states.
  • #1
scg4d
6
0
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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  • #2
"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.
 
  • #3
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.
 
  • #4
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?
 
  • #5
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...?
 
  • #6
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).
 

What is the Exponential Distribution Problem?

The Exponential Distribution Problem is a statistical problem that involves finding the probability distribution of the time between events in a Poisson process, where events occur continuously and independently at a constant average rate.

What is the formula for the Exponential Distribution?

The formula for the Exponential Distribution is f(x) = λe^(-λx), where λ is the rate parameter and x is the time between events.

What are the properties of the Exponential Distribution?

The Exponential Distribution has two main properties: memorylessness and lack of memory. This means that the probability of an event occurring in the future is not affected by how long we have already waited for the event to occur.

How is the Exponential Distribution used in real life?

The Exponential Distribution is commonly used in various fields, such as finance, engineering, and biology. It can be used to model the time between customer arrivals, the lifespan of a machine, or the time between radioactive decay of atoms.

What are some common misconceptions about the Exponential Distribution?

One common misconception about the Exponential Distribution is that it can only be used for continuous data. However, it can also be used for discrete data if the data is converted into continuous time intervals. Another misconception is that the Exponential Distribution can only model events that occur at a constant rate, when in fact it can also be used for events with changing rates, as long as the average rate is known.

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