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  #1  
Old 05-07-2019, 10:19 AM
actuaryxjh actuaryxjh is offline
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Default 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.
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Old 05-07-2019, 11:24 AM
Actuarialsuck Actuarialsuck is offline
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Are you asking for something like this? https://www.quora.com/What-are-examp...s-in-real-life
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Old 05-07-2019, 11:38 AM
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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?
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Old 05-07-2019, 01:02 PM
tyder21 tyder21 is offline
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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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Old 05-07-2019, 06:18 PM
actuaryxjh actuaryxjh is offline
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Originally Posted by tyder21 View Post
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?
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Old 05-07-2019, 06:21 PM
actuaryxjh actuaryxjh is offline
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Originally Posted by Gandalf View Post
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?
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Old 05-07-2019, 06:21 PM
actuaryxjh actuaryxjh is offline
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Are you asking for something like this? https://www.quora.com/What-are-examp...s-in-real-life
Nope, that's not what I was asking, but thank you for your reply~
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Old 05-07-2019, 06:39 PM
Academic Actuary Academic Actuary is offline
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Originally Posted by actuaryxjh View Post
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.
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Old 05-07-2019, 07:47 PM
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Quote:
Originally Posted by actuaryxjh View Post
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.
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Old 05-08-2019, 09:20 PM
actuaryxjh actuaryxjh is offline
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Originally Posted by Academic Actuary View Post
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?
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