Thread: An exponential random variable question

Hello to all,

let's say i have 2 i.i.d exponential random variable x1 and x2 with parameter m.

i want to find out the probability that x.1 is equal to x.2 : P(x.1 = x.2)

so what i did is to say that by conditioning on x.2 i get : (1) P(x.1 = x.2) = P(x.1 = k | x.2 = k)*P(x.2 = k) and i will sum over all possible values of k (by integral) [0,Infinity].

because we have 2 i.i.d variables i can say that : (2) P(x.1 = k | x.2 = k) = P(x.1 = k)

so by inserting equation (1) -> (2) i get (3) P(x.1 = x.2) = P(x.1 = k)*P(x.2 = k) sum over all possible values of k (by integral) [0,Infinity].

now all i have to do is to place P(x.1 = k) and P(x.2 = k) into (3) and i know that P(x = k) = f(k) = m*e^(-m*k) for both variables because they are i.i.d.

so i get equation no. (4) P(x.1 = x.2) = (m^2)*e^(-2*m*k) sum over all possible values of k (by integral) [0,Infinity].

if i solve this integral i get (5) P(x.1 = x.2) = m/2

everything seems o.k by the way i developed this answer but what if m=5 ? the probability can't be 2.5 but m surely can be 5...

so where did i get it wrong ? in the way or in the understanding of the answer i got ?

thanks in advance...

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Hey DudeM.

You won't be able to check an exact inequality since it is a continuous random variable, but what you can do is check an interval.

So basically you want to find whether the difference lies in a particular interval of [-e,e] for some small e > 0.

You can find the sum of two variables distribution by using convolution.

From this:

List of convolutions of probability distributions - Wikipedia, the free encyclopedia

we can see that the distribution will be a gamma however this is only for summing random variables. Yheoutiou will need to find the PDF of the -X where X is exponential and then use the convolution theorem to get the PDF.

Once you have the PDF then just calculate the probability in [-e,e]

"You won't be able to check an exact inequality since it is a continuous random variable..."

what do you mean by that? by using the exponential variable pdf i can find the probability that x.1 = k , since P(x.1 = k) = f.x(k) -> the pdf at point k.

by the way , why did you say i was summing random variable.... maybe i do x.1 - x.2, ( from P(x.1 = x.2) ) thats alot different...

What I mean is that for any continuous random variable P(X = x) is 0.

This is not the same as the probability density function which has a specific non-zero, valid probability value.

The actual probability of getting one observation as opposed to a range of probabilities is zero.