What's the probability of drawing a number from [0,1]?
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The question that I have is more or less equal to this one. However, it is slightly different, and since I wasn't able to completely follow the answer, I decided to post a new question:
So I draw two numbers from $[0,1]$ and want their sum to be greater or equal to $0.5$.
It is not stated in the question what should be the probability distribution. I assume the random variables are independent and identically distributed.
What I didn't understand from the top-rated answer to the question given in the link is the transition of the second last to the last integral:
$$int_0^1f(x)left(int_{1-x}^1f(y),mathrm dyright),mathrm dx=int_0^1f(x)big(1-F(1-x)big),mathrm dx$$
Anyways, I am pretty sure for my problem I just have to change the lower bound of the second last integral from $1-x$ to $0.5 - x$. But what would be the last transition then? I know the integral over the probability density function $f(y)$ equals the cumulative distribution function $F(y)$. But that's about where my statistical knowledge leaves me.
I appreciate any help, thx in advance!
probability statistics
edited Nov 25 at 13:44
drhab
95.2k543126
asked Nov 25 at 13:33
user503842
103
add a comment |
up vote
0
down vote
favorite
The question that I have is more or less equal to this one. However, it is slightly different, and since I wasn't able to completely follow the answer, I decided to post a new question:
So I draw two numbers from $[0,1]$ and want their sum to be greater or equal to $0.5$.
It is not stated in the question what should be the probability distribution. I assume the random variables are independent and identically distributed.
What I didn't understand from the top-rated answer to the question given in the link is the transition of the second last to the last integral:
$$int_0^1f(x)left(int_{1-x}^1f(y),mathrm dyright),mathrm dx=int_0^1f(x)big(1-F(1-x)big),mathrm dx$$
Anyways, I am pretty sure for my problem I just have to change the lower bound of the second last integral from $1-x$ to $0.5 - x$. But what would be the last transition then? I know the integral over the probability density function $f(y)$ equals the cumulative distribution function $F(y)$. But that's about where my statistical knowledge leaves me.
I appreciate any help, thx in advance!
probability statistics
edited Nov 25 at 13:44
drhab
95.2k543126
asked Nov 25 at 13:33
user503842
103
Do you understand that here $int_{1-x}^1f(y)dy=F(1)-F(1-x)=1-F(1-x)$?
– drhab
Nov 25 at 13:47
yeah, I think so. The first equation is just a property of the cumulative distribution function:int_{a}^{b} f(x) dx = F(b) - F(a). And F(b) = 1. So that explains this equation. But how do I continue with my problem? I would say I end up with the formulaint_{0}^{1} f(x) (1-F(0.5-x)) dx, since y = 0.5-x. And what if I now assume a uniform distribution? the probability density f(x) = 1/(b-a) in the interval [a,b]. ...
– user503842
Nov 25 at 14:10
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The question that I have is more or less equal to this one. However, it is slightly different, and since I wasn't able to completely follow the answer, I decided to post a new question:
So I draw two numbers from $[0,1]$ and want their sum to be greater or equal to $0.5$.
It is not stated in the question what should be the probability distribution. I assume the random variables are independent and identically distributed.
What I didn't understand from the top-rated answer to the question given in the link is the transition of the second last to the last integral:
$$int_0^1f(x)left(int_{1-x}^1f(y),mathrm dyright),mathrm dx=int_0^1f(x)big(1-F(1-x)big),mathrm dx$$
Anyways, I am pretty sure for my problem I just have to change the lower bound of the second last integral from $1-x$ to $0.5 - x$. But what would be the last transition then? I know the integral over the probability density function $f(y)$ equals the cumulative distribution function $F(y)$. But that's about where my statistical knowledge leaves me.
I appreciate any help, thx in advance!
probability statistics
edited Nov 25 at 13:44
drhab
95.2k543126
asked Nov 25 at 13:33
user503842
103
The question that I have is more or less equal to this one. However, it is slightly different, and since I wasn't able to completely follow the answer, I decided to post a new question:
So I draw two numbers from $[0,1]$ and want their sum to be greater or equal to $0.5$.
It is not stated in the question what should be the probability distribution. I assume the random variables are independent and identically distributed.
What I didn't understand from the top-rated answer to the question given in the link is the transition of the second last to the last integral:
$$int_0^1f(x)left(int_{1-x}^1f(y),mathrm dyright),mathrm dx=int_0^1f(x)big(1-F(1-x)big),mathrm dx$$
Anyways, I am pretty sure for my problem I just have to change the lower bound of the second last integral from $1-x$ to $0.5 - x$. But what would be the last transition then? I know the integral over the probability density function $f(y)$ equals the cumulative distribution function $F(y)$. But that's about where my statistical knowledge leaves me.
I appreciate any help, thx in advance!
probability statistics
probability statistics
edited Nov 25 at 13:44
drhab
95.2k543126
asked Nov 25 at 13:33
user503842
103
edited Nov 25 at 13:44
drhab
95.2k543126
asked Nov 25 at 13:33
user503842
103
edited Nov 25 at 13:44
drhab
95.2k543126
edited Nov 25 at 13:44
drhab
95.2k543126
edited Nov 25 at 13:44
drhab
95.2k543126
95.2k543126
asked Nov 25 at 13:33
user503842
103
asked Nov 25 at 13:33
user503842
103
asked Nov 25 at 13:33
user503842
103
103
Do you understand that here $int_{1-x}^1f(y)dy=F(1)-F(1-x)=1-F(1-x)$?
– drhab
Nov 25 at 13:47
yeah, I think so. The first equation is just a property of the cumulative distribution function:int_{a}^{b} f(x) dx = F(b) - F(a). And F(b) = 1. So that explains this equation. But how do I continue with my problem? I would say I end up with the formulaint_{0}^{1} f(x) (1-F(0.5-x)) dx, since y = 0.5-x. And what if I now assume a uniform distribution? the probability density f(x) = 1/(b-a) in the interval [a,b]. ...
– user503842
Nov 25 at 14:10
add a comment |
Do you understand that here $int_{1-x}^1f(y)dy=F(1)-F(1-x)=1-F(1-x)$?
– drhab
Nov 25 at 13:47
yeah, I think so. The first equation is just a property of the cumulative distribution function:int_{a}^{b} f(x) dx = F(b) - F(a). And F(b) = 1. So that explains this equation. But how do I continue with my problem? I would say I end up with the formulaint_{0}^{1} f(x) (1-F(0.5-x)) dx, since y = 0.5-x. And what if I now assume a uniform distribution? the probability density f(x) = 1/(b-a) in the interval [a,b]. ...
– user503842
Nov 25 at 14:10
Do you understand that here $int_{1-x}^1f(y)dy=F(1)-F(1-x)=1-F(1-x)$?
– drhab
Nov 25 at 13:47
Do you understand that here $int_{1-x}^1f(y)dy=F(1)-F(1-x)=1-F(1-x)$?
– drhab
Nov 25 at 13:47
yeah, I think so. The first equation is just a property of the cumulative distribution function:
int_{a}^{b} f(x) dx = F(b) - F(a). And F(b) = 1. So that explains this equation. But how do I continue with my problem? I would say I end up with the formula int_{0}^{1} f(x) (1-F(0.5-x)) dx, since y = 0.5-x. And what if I now assume a uniform distribution? the probability density f(x) = 1/(b-a) in the interval [a,b]. ...– user503842
Nov 25 at 14:10
yeah, I think so. The first equation is just a property of the cumulative distribution function:
int_{a}^{b} f(x) dx = F(b) - F(a). And F(b) = 1. So that explains this equation. But how do I continue with my problem? I would say I end up with the formula int_{0}^{1} f(x) (1-F(0.5-x)) dx, since y = 0.5-x. And what if I now assume a uniform distribution? the probability density f(x) = 1/(b-a) in the interval [a,b]. ...– user503842
Nov 25 at 14:10
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
It is more convenient here to use $P(X+Ygeq0.5)=1-P(X+Y<0.5)$ and calculate $P(X+Y<0.5)$.
If $[x+y<0.5]$ denotes the function $mathbb R^2tomathbb R$ that takes value $1$ if $x+y<0.5$ and takes value $0$ otherwise then:$$P(X+Y<0.5)=mathbb E[X+Y<0.5]=int_0^1int_0^1[x+y<0.5]f(x)f(y)dydx=$$$$int_0^{0.5}int_0^{0.5-x}f(x)f(y)dydx=int_0^{0.5}f(x)left(int_0^{0.5-x}f(y)dyright)dx=int_0^{0.5}f(x)F(0.5-x)dx$$so that$$P(X+Ygeq0.5)=1-int_0^{0.5}f(x)F(0.5-x)dx$$
Observe that: $$int_0^{0.5-x}f(y)dy=F(0.5-x)-F(0)=F(0.5-x)$$where the first equality is purely based on the definition of a PDF.
answered Nov 25 at 14:06
drhab
95.2k543126
ahh, allright, thx so much drhab!!! Just to be sure: Say I assume a uniform distribution: This would yield for P(X + Y < 0.5):int_{0}^{0.5} (1*(0.5-x)) dxwhich eventually solves to 0.00125. Is that correct?
– user503842
Nov 25 at 14:22
By uniform distribution: $int_0^{0.5}(0.5-x)dx=[0.5x-0.5x^2]_0^{0.5}=[0.5^2-0.5^3]-0=0.125$
– drhab
Nov 25 at 15:37
allright. thx so much!
– user503842
Nov 25 at 17:41
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
It is more convenient here to use $P(X+Ygeq0.5)=1-P(X+Y<0.5)$ and calculate $P(X+Y<0.5)$.
If $[x+y<0.5]$ denotes the function $mathbb R^2tomathbb R$ that takes value $1$ if $x+y<0.5$ and takes value $0$ otherwise then:$$P(X+Y<0.5)=mathbb E[X+Y<0.5]=int_0^1int_0^1[x+y<0.5]f(x)f(y)dydx=$$$$int_0^{0.5}int_0^{0.5-x}f(x)f(y)dydx=int_0^{0.5}f(x)left(int_0^{0.5-x}f(y)dyright)dx=int_0^{0.5}f(x)F(0.5-x)dx$$so that$$P(X+Ygeq0.5)=1-int_0^{0.5}f(x)F(0.5-x)dx$$
Observe that: $$int_0^{0.5-x}f(y)dy=F(0.5-x)-F(0)=F(0.5-x)$$where the first equality is purely based on the definition of a PDF.
answered Nov 25 at 14:06
drhab
95.2k543126
ahh, allright, thx so much drhab!!! Just to be sure: Say I assume a uniform distribution: This would yield for P(X + Y < 0.5):int_{0}^{0.5} (1*(0.5-x)) dxwhich eventually solves to 0.00125. Is that correct?
– user503842
Nov 25 at 14:22
By uniform distribution: $int_0^{0.5}(0.5-x)dx=[0.5x-0.5x^2]_0^{0.5}=[0.5^2-0.5^3]-0=0.125$
– drhab
Nov 25 at 15:37
allright. thx so much!
– user503842
Nov 25 at 17:41
add a comment |
up vote
0
down vote
accepted
It is more convenient here to use $P(X+Ygeq0.5)=1-P(X+Y<0.5)$ and calculate $P(X+Y<0.5)$.
If $[x+y<0.5]$ denotes the function $mathbb R^2tomathbb R$ that takes value $1$ if $x+y<0.5$ and takes value $0$ otherwise then:$$P(X+Y<0.5)=mathbb E[X+Y<0.5]=int_0^1int_0^1[x+y<0.5]f(x)f(y)dydx=$$$$int_0^{0.5}int_0^{0.5-x}f(x)f(y)dydx=int_0^{0.5}f(x)left(int_0^{0.5-x}f(y)dyright)dx=int_0^{0.5}f(x)F(0.5-x)dx$$so that$$P(X+Ygeq0.5)=1-int_0^{0.5}f(x)F(0.5-x)dx$$
Observe that: $$int_0^{0.5-x}f(y)dy=F(0.5-x)-F(0)=F(0.5-x)$$where the first equality is purely based on the definition of a PDF.
answered Nov 25 at 14:06
drhab
95.2k543126
ahh, allright, thx so much drhab!!! Just to be sure: Say I assume a uniform distribution: This would yield for P(X + Y < 0.5):int_{0}^{0.5} (1*(0.5-x)) dxwhich eventually solves to 0.00125. Is that correct?
– user503842
Nov 25 at 14:22
By uniform distribution: $int_0^{0.5}(0.5-x)dx=[0.5x-0.5x^2]_0^{0.5}=[0.5^2-0.5^3]-0=0.125$
– drhab
Nov 25 at 15:37
allright. thx so much!
– user503842
Nov 25 at 17:41
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
It is more convenient here to use $P(X+Ygeq0.5)=1-P(X+Y<0.5)$ and calculate $P(X+Y<0.5)$.
If $[x+y<0.5]$ denotes the function $mathbb R^2tomathbb R$ that takes value $1$ if $x+y<0.5$ and takes value $0$ otherwise then:$$P(X+Y<0.5)=mathbb E[X+Y<0.5]=int_0^1int_0^1[x+y<0.5]f(x)f(y)dydx=$$$$int_0^{0.5}int_0^{0.5-x}f(x)f(y)dydx=int_0^{0.5}f(x)left(int_0^{0.5-x}f(y)dyright)dx=int_0^{0.5}f(x)F(0.5-x)dx$$so that$$P(X+Ygeq0.5)=1-int_0^{0.5}f(x)F(0.5-x)dx$$
Observe that: $$int_0^{0.5-x}f(y)dy=F(0.5-x)-F(0)=F(0.5-x)$$where the first equality is purely based on the definition of a PDF.
answered Nov 25 at 14:06
drhab
95.2k543126
It is more convenient here to use $P(X+Ygeq0.5)=1-P(X+Y<0.5)$ and calculate $P(X+Y<0.5)$.
If $[x+y<0.5]$ denotes the function $mathbb R^2tomathbb R$ that takes value $1$ if $x+y<0.5$ and takes value $0$ otherwise then:$$P(X+Y<0.5)=mathbb E[X+Y<0.5]=int_0^1int_0^1[x+y<0.5]f(x)f(y)dydx=$$$$int_0^{0.5}int_0^{0.5-x}f(x)f(y)dydx=int_0^{0.5}f(x)left(int_0^{0.5-x}f(y)dyright)dx=int_0^{0.5}f(x)F(0.5-x)dx$$so that$$P(X+Ygeq0.5)=1-int_0^{0.5}f(x)F(0.5-x)dx$$
Observe that: $$int_0^{0.5-x}f(y)dy=F(0.5-x)-F(0)=F(0.5-x)$$where the first equality is purely based on the definition of a PDF.
answered Nov 25 at 14:06
drhab
95.2k543126
answered Nov 25 at 14:06
drhab
95.2k543126
answered Nov 25 at 14:06
drhab
95.2k543126
answered Nov 25 at 14:06
drhab
95.2k543126
95.2k543126
ahh, allright, thx so much drhab!!! Just to be sure: Say I assume a uniform distribution: This would yield for P(X + Y < 0.5):int_{0}^{0.5} (1*(0.5-x)) dxwhich eventually solves to 0.00125. Is that correct?
– user503842
Nov 25 at 14:22
By uniform distribution: $int_0^{0.5}(0.5-x)dx=[0.5x-0.5x^2]_0^{0.5}=[0.5^2-0.5^3]-0=0.125$
– drhab
Nov 25 at 15:37
allright. thx so much!
– user503842
Nov 25 at 17:41
add a comment |
ahh, allright, thx so much drhab!!! Just to be sure: Say I assume a uniform distribution: This would yield for P(X + Y < 0.5):int_{0}^{0.5} (1*(0.5-x)) dxwhich eventually solves to 0.00125. Is that correct?
– user503842
Nov 25 at 14:22
By uniform distribution: $int_0^{0.5}(0.5-x)dx=[0.5x-0.5x^2]_0^{0.5}=[0.5^2-0.5^3]-0=0.125$
– drhab
Nov 25 at 15:37
allright. thx so much!
– user503842
Nov 25 at 17:41
ahh, allright, thx so much drhab!!! Just to be sure: Say I assume a uniform distribution: This would yield for P(X + Y < 0.5):
int_{0}^{0.5} (1*(0.5-x)) dx which eventually solves to 0.00125. Is that correct?– user503842
Nov 25 at 14:22
ahh, allright, thx so much drhab!!! Just to be sure: Say I assume a uniform distribution: This would yield for P(X + Y < 0.5):
int_{0}^{0.5} (1*(0.5-x)) dx which eventually solves to 0.00125. Is that correct?– user503842
Nov 25 at 14:22
By uniform distribution: $int_0^{0.5}(0.5-x)dx=[0.5x-0.5x^2]_0^{0.5}=[0.5^2-0.5^3]-0=0.125$
– drhab
Nov 25 at 15:37
By uniform distribution: $int_0^{0.5}(0.5-x)dx=[0.5x-0.5x^2]_0^{0.5}=[0.5^2-0.5^3]-0=0.125$
– drhab
Nov 25 at 15:37
allright. thx so much!
– user503842
Nov 25 at 17:41
allright. thx so much!
– user503842
Nov 25 at 17:41
add a comment |
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Do you understand that here $int_{1-x}^1f(y)dy=F(1)-F(1-x)=1-F(1-x)$?
– drhab
Nov 25 at 13:47
yeah, I think so. The first equation is just a property of the cumulative distribution function:
int_{a}^{b} f(x) dx = F(b) - F(a). And F(b) = 1. So that explains this equation. But how do I continue with my problem? I would say I end up with the formulaint_{0}^{1} f(x) (1-F(0.5-x)) dx, since y = 0.5-x. And what if I now assume a uniform distribution? the probability density f(x) = 1/(b-a) in the interval [a,b]. ...– user503842
Nov 25 at 14:10