Actuarial Outpost example of continuous probability
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 Probability Old Exam P Forum

#1
05-07-2019, 10:19 AM
 actuaryxjh SOA Join Date: Aug 2018 College: Rutgers University Posts: 20
example of continuous probability

Can someone please give me a real-life example of this probability so that I can understand more thoroughly?

f(x) = 2x for 0<x<1, and 0 elsewhere.
#2
05-07-2019, 11:24 AM
 Actuarialsuck Member Join Date: Sep 2007 Posts: 6,147

Are you asking for something like this? https://www.quora.com/What-are-examp...s-in-real-life
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#3
05-07-2019, 11:38 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 31,286

I think this example works. A point is chosen from the unit square, with lower left corner (0,0) and upper right corner (1,1). What is the probability that both coordinates of the point are less than x?
#4
05-07-2019, 01:02 PM
 tyder21 Member CAS Join Date: May 2012 Location: NYC Studying for NOTHING Posts: 132

If you draw 2 random numbers between 0 and 1, the PDF of the larger of the two numbers is f(x) = 2x.

Fun Fact: This can actually be generalized more. If you draw n random numbers between 0 and 1, the PDF of the largest number is f(x) = n * x^(n - 1). Expected value of the largest number is n / (n + 1).
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#5
05-07-2019, 06:18 PM
 actuaryxjh SOA Join Date: Aug 2018 College: Rutgers University Posts: 20

Quote:
 Originally Posted by tyder21 If you draw 2 random numbers between 0 and 1, the PDF of the larger of the two numbers is f(x) = 2x. Fun Fact: This can actually be generalized more. If you draw n random numbers between 0 and 1, the PDF of the largest number is f(x) = n * x^(n - 1). Expected value of the largest number is n / (n + 1).
I am sorry I don't quite get what you mean. So by what you are saying, if you randomly pick two numbers from 0 to 1, then the probability that the larger number is 0.5 equals to 1?
#6
05-07-2019, 06:21 PM
 actuaryxjh SOA Join Date: Aug 2018 College: Rutgers University Posts: 20

Quote:
 Originally Posted by Gandalf I think this example works. A point is chosen from the unit square, with lower left corner (0,0) and upper right corner (1,1). What is the probability that both coordinates of the point are less than x?
Let's consider this, if I let x equals to 0.1, then f(x) = 2x = 0.2, but in fact, both cooridinate less than 0.1 is confined in 0.1*0.1 square, which I think the probability is （0.1*0.1）/（1*1） = 0.01？
#7
05-07-2019, 06:21 PM
 actuaryxjh SOA Join Date: Aug 2018 College: Rutgers University Posts: 20

Quote:
 Originally Posted by Actuarialsuck Are you asking for something like this? https://www.quora.com/What-are-examp...s-in-real-life
#8
05-07-2019, 06:39 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 8,925

Quote:
 Originally Posted by actuaryxjh I am sorry I don't quite get what you mean. So by what you are saying, if you randomly pick two numbers from 0 to 1, then the probability that the larger number is 0.5 equals to 1?
With continuous random variables single points have zero probability. If f(x) = 2x, then F(x) = x^2, so the probability the larger of the two random numbers is less than .5 is .25. Let's say we want the probability that the larger of the two is between .4 and .6. We would integrate the pdf over that interval.
#9
05-07-2019, 07:47 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 31,286

Quote:
 Originally Posted by actuaryxjh Let's consider this, if I let x equals to 0.1, then f(x) = 2x = 0.2, but in fact, both cooridinate less than 0.1 is confined in 0.1*0.1 square, which I think the probability is （0.1*0.1）/（1*1） = 0.01？
Yes, the probability is 0.01. As pointed out by Academic Actuary later in the thread, probabilities are integrals of the density function, not the density function. The integral of the density function from 0 to .1 is .01.
#10
05-08-2019, 09:20 PM
 actuaryxjh SOA Join Date: Aug 2018 College: Rutgers University Posts: 20

Quote:
 Originally Posted by Academic Actuary With continuous random variables single points have zero probability. If f(x) = 2x, then F(x) = x^2, so the probability the larger of the two random numbers is less than .5 is .25. Let's say we want the probability that the larger of the two is between .4 and .6. We would integrate the pdf over that interval.
Oh, you are right, my mistake~

The question is, how can we show that the pdf is actually f(x) = 2x in this case?