uniform random variables

aledg97

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X and Y are two independent uniformly distributed random variables in [0; 1]. Then

P(Y-X<=1/2) is ??

Now from a graphical point of view, I plot the unit side square [0,1], and the line y= x+1/2, since I want to find the area which is below the triangle, I do 1- area(triangle) = 1- 1/2*1/2*1/2 = 7/8
and that's the correct answer

But I want to find that probability applying a general method, using the density function and making a double integral. They are independent

so I have to
\[\int_{0}^{1}\int_{0}^{1/2+x} fxfy dy dx\]=\[\int_{0}^{1}\int_{0}^{1/2+x} 1 dy dx\]
because fxfy =fx*fy= 1*1=1

But I get 1, so I make mistakes in defining the extrema
 
X and Y are two independent uniformly distributed random variables in [0; 1]. Then

P(Y-X<=1/2) is ??

Now from a graphical point of view, I plot the unit side square [0,1], and the line y= x+1/2, since I want to find the area which is below the triangle, I do 1- area(triangle) = 1- 1/2*1/2*1/2 = 7/8
and that's the correct answer

But I want to find that probability applying a general method, using the density function and making a double integral. They are independent

so I have to
\[\int_{0}^{1}\int_{0}^{1/2+x} fxfy dy dx\]=\[\int_{0}^{1}\int_{0}^{1/2+x} 1 dy dx\]
because fxfy =fx*fy= 1*1=1

But I get 1, so I make mistakes in defining the extrema
Check those limits. \(\displaystyle y\in[0,x+1/2]\) ONLY for \(\displaystyle x\in[0,1/2]\). Rethink for \(\displaystyle x\in[1/2,1]\)

You would do well to reverse the order of integration and paint the small piece. It might make your life easier.
 
Check those limits. \(\displaystyle y\in[0,x+1/2]\) ONLY for \(\displaystyle x\in[0,1/2]\). Rethink for \(\displaystyle x\in[1/2,1]\)

You would do well to reverse the order of integration and paint the small piece. It might make your life easier.


OK, thanks. From drawing the figure I understand what you said, so I could split the integral in 2. First when x is between 0 and 1/2, and then when it is between 1/2 and 1.

But what do you mean by changing the order of integration? Is there a way to make only one integral without splitting?
 
OK, thanks. From drawing the figure I understand what you said, so I could split the integral in 2. First when x is between 0 and 1/2, and then when it is between 1/2 and 1.

But what do you mean by changing the order of integration? Is there a way to make only one integral without splitting?


\(\displaystyle \int\limits_{1/2}^{1}\int\limits_{0}^{y-1/2} dx\;dy\) -- Just the little piece. Subtract from 1.
 
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