A Mixed Distribution Example

MIT 6.041SC Probabilistic Systems Analysis and Applied Probability, Fall 2013
Instructor: Jagdish Ramakrishnan

License: Creative Commons BY-NC-SA

[sakichan]

This solution is far better than the books one, Amazing!!

[thishandledoesntexist]

We are not bothered if the taxi arrives after 5 minutes, why? because Al will take the bus instead which we know for certain arrives in 5 minutes. The situation that the taxi arrives after 5 minutes does not factor into the calculations of the expected waiting time.

[annawilson]

before I saw the video, the noise made me think he is on the plane...

[laodrofotic]

this exercise is right but needs a small detail that I got or maybe I am wrong..?. the probability that the next taxi arrives is between 0 and 5 but the video didnt explain why correctly i think?

[milosnovakovic]

Ok so basically, I think he was wrong with calculating the EX. Why do I think that? Well let me explain:


-So first of all, this is a mixed distribution, therefore that means it has properties similar with discrete and continuous random variables (RV), so it has something combined with PMF (discrete property) and PDF (continuous property). In this case, the probability that X = 0 is 2/3, or for short notation P(X=0) = 2/3, so that implies that
P(X>0) = 1 - P(X=0) = 1/3.


-Now we look furthermore. We move to other branch of the tree where X>0 (we multiply by P(X>0)), so we are asking next question :
What is the probability that P(X = 5) = P(Al drove by a bus, and not by a taxi)?
Well that means that the taxi didn't come in less than 5, so that means, that
P(X=5) = P(X>0) * P(5 <= <= 10 | X>0) = 1/3* integral from 5 till 10 (1/10)*dx = 1/3 * (10-5)/10 = 5/30 = 1/6


-So we get that P(X=0) = 2/3 and that P(X=5) = 1/6. And that is the discrete part of X. Now to the continuous part of random variable X:
What is PDF of random variable X f_X(x), when x is between 0 and 5?
So, PDF of random variable X is: f_X(x) = P(X>0) * = 1/3 * 1/10 = 1/30, where x is continueous random variable from 0 till 5.


-So lets conclude: P(X=0) = 2/3 and f_X(x) = 1/30 (where x is between 0 and 5) and P(X=5) = 1/6. Then, we need to calculate the EX, where we have the discrete part (the sumation) and the continuous part (the integration):


-EX = sum (x * P(X = x)) + integral(x*f_X(x))*dx =
= 0*P(X=0) + 5*P(X=5) + integral_from_0_to_5 (x * f_X(x))*dx =
= 0*2/3 + 5*1/6 + integral_from_0_to_5 (x*1/30)*dx =
= 0 + 5/6 + 5/30 =
= 0 + 5/6 + 1/6 = 1


-So there you go! Hope it helps! <3 :))

[worstyearever]

why is the expected value of the density 5/2 and not 1/2, i thought for pdf you are looking for the area under the curve -- or whatever, so wouldnt it be height (1/5) x base (5/2)? 

[jackypaulcukjati]

why is the E[X|B2] = 5/2
Is it not the integral from 0 -> 5 0.1x dx = 1.25 because the taxi is a U [0, 10]

[manishsahoo]

Approach seems wrong although answer is correct

[hanusatjaldrafltti]

Didn't watch because of the bad audio