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Let X and Y be two random variables such that the vector (X,Y) is uniformly distributed over the region R = {(x,y)∈ℝ^2 : 0

0

$begingroup$

Im having a hard time beginning because I don't know how to find the distribution function, and thus don't understand how to find the density function. I drew the graph and found the triangle. What do I do from there on?

probability probability-theory statistics random-variables

share|cite|improve this question

asked Mar 27 at 3:08

Malaikatu KargboMalaikatu Kargbo

82

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A friendly advice: title is NOT the first sentence of your question. In particular, see the last bullet: the question post should be comprehensible without the title, even though one should make good use of the title to provide extra info.
  $endgroup$
  – Lee David Chung Lin
  Mar 27 at 3:35
  

  
  
  
  

add a comment | 

0

$begingroup$

Im having a hard time beginning because I don't know how to find the distribution function, and thus don't understand how to find the density function. I drew the graph and found the triangle. What do I do from there on?

probability probability-theory statistics random-variables

share|cite|improve this question

asked Mar 27 at 3:08

Malaikatu KargboMalaikatu Kargbo

82

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A friendly advice: title is NOT the first sentence of your question. In particular, see the last bullet: the question post should be comprehensible without the title, even though one should make good use of the title to provide extra info.
  $endgroup$
  – Lee David Chung Lin
  Mar 27 at 3:35
  

  
  
  
  

add a comment | 

$begingroup$

Im having a hard time beginning because I don't know how to find the distribution function, and thus don't understand how to find the density function. I drew the graph and found the triangle. What do I do from there on?

probability probability-theory statistics random-variables

share|cite|improve this question

asked Mar 27 at 3:08

Malaikatu KargboMalaikatu Kargbo

82

$endgroup$

share|cite|improve this question

asked Mar 27 at 3:08

Malaikatu KargboMalaikatu Kargbo

82

share|cite|improve this question

asked Mar 27 at 3:08

Malaikatu KargboMalaikatu Kargbo

82

asked Mar 27 at 3:08

Malaikatu KargboMalaikatu Kargbo

82

asked Mar 27 at 3:08

Malaikatu KargboMalaikatu Kargbo

82

1 Answer
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1

$begingroup$

Let X and Y be two random variables such that the vector (X,Y) is uniformly distributed over the region $R = (x,y)inBbb R^2 : 0<x<y<1$ . Find P(X+Y < 1)

I'm having a hard time beginning because I don't know how to find the distribution function, and thus don't understand how to find the density function.

If you wish to use it, you should find the probability density function first.

You don't strictly need it, but anyway...

I drew the graph and found the triangle.

Indeed, the region $R$ is a triangle of area $1/2$. The random points are uniformly distributed over this.

The probability density function is therefore : $f_X,Y(x,y) = 2cdotmathbf 1_0<x<y<1$ $$f_X,Y(x,y) = begincases2&:& 0<x<y<1\0&:&textsfotherwiseendcases$$

What do I do from there on?

Draw the line $x+y=1$ on your graph. Notice how it divides the triangle. The ratio of areas will give the probability.

$$mathsf P(X+Y<1) =dfraclvert Rcap(x,y)inBbb R^2:x+y<1rvertlvert R rvert$$

Or you might do it the long way. If you really want to:

$$beginalignmathsf P(X+Ylt 1)&=int_0^1int_0^min(y,1-y)2~mathrm d x~mathrm d y\[1ex]&=int_0^1/2int_0^y 2mathsf d xmathsf d y+int_1/2^1int_0^1-y 2mathsf d xmathsf d y\[1ex]&~~vdotsendalign$$

share|cite|improve this answer

answered Mar 27 at 3:42

Graham KempGraham Kemp

88k43579

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add a comment | 

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1

$begingroup$

Let X and Y be two random variables such that the vector (X,Y) is uniformly distributed over the region $R = (x,y)inBbb R^2 : 0<x<y<1$ . Find P(X+Y < 1)

I'm having a hard time beginning because I don't know how to find the distribution function, and thus don't understand how to find the density function.

If you wish to use it, you should find the probability density function first.

You don't strictly need it, but anyway...

I drew the graph and found the triangle.

Indeed, the region $R$ is a triangle of area $1/2$. The random points are uniformly distributed over this.

The probability density function is therefore : $f_X,Y(x,y) = 2cdotmathbf 1_0<x<y<1$ $$f_X,Y(x,y) = begincases2&:& 0<x<y<1\0&:&textsfotherwiseendcases$$

What do I do from there on?

Draw the line $x+y=1$ on your graph. Notice how it divides the triangle. The ratio of areas will give the probability.

$$mathsf P(X+Y<1) =dfraclvert Rcap(x,y)inBbb R^2:x+y<1rvertlvert R rvert$$

Or you might do it the long way. If you really want to:

$$beginalignmathsf P(X+Ylt 1)&=int_0^1int_0^min(y,1-y)2~mathrm d x~mathrm d y\[1ex]&=int_0^1/2int_0^y 2mathsf d xmathsf d y+int_1/2^1int_0^1-y 2mathsf d xmathsf d y\[1ex]&~~vdotsendalign$$

share|cite|improve this answer

answered Mar 27 at 3:42

Graham KempGraham Kemp

88k43579

$endgroup$

add a comment | 

1

$begingroup$

Let X and Y be two random variables such that the vector (X,Y) is uniformly distributed over the region $R = (x,y)inBbb R^2 : 0<x<y<1$ . Find P(X+Y < 1)

I'm having a hard time beginning because I don't know how to find the distribution function, and thus don't understand how to find the density function.

If you wish to use it, you should find the probability density function first.

You don't strictly need it, but anyway...

I drew the graph and found the triangle.

Indeed, the region $R$ is a triangle of area $1/2$. The random points are uniformly distributed over this.

The probability density function is therefore : $f_X,Y(x,y) = 2cdotmathbf 1_0<x<y<1$ $$f_X,Y(x,y) = begincases2&:& 0<x<y<1\0&:&textsfotherwiseendcases$$

What do I do from there on?

Draw the line $x+y=1$ on your graph. Notice how it divides the triangle. The ratio of areas will give the probability.

$$mathsf P(X+Y<1) =dfraclvert Rcap(x,y)inBbb R^2:x+y<1rvertlvert R rvert$$

Or you might do it the long way. If you really want to:

$$beginalignmathsf P(X+Ylt 1)&=int_0^1int_0^min(y,1-y)2~mathrm d x~mathrm d y\[1ex]&=int_0^1/2int_0^y 2mathsf d xmathsf d y+int_1/2^1int_0^1-y 2mathsf d xmathsf d y\[1ex]&~~vdotsendalign$$

share|cite|improve this answer

answered Mar 27 at 3:42

Graham KempGraham Kemp

88k43579

$endgroup$

add a comment | 

$begingroup$

Let X and Y be two random variables such that the vector (X,Y) is uniformly distributed over the region $R = (x,y)inBbb R^2 : 0<x<y<1$ . Find P(X+Y < 1)

I'm having a hard time beginning because I don't know how to find the distribution function, and thus don't understand how to find the density function.

If you wish to use it, you should find the probability density function first.

You don't strictly need it, but anyway...

I drew the graph and found the triangle.

Indeed, the region $R$ is a triangle of area $1/2$. The random points are uniformly distributed over this.

The probability density function is therefore : $f_X,Y(x,y) = 2cdotmathbf 1_0<x<y<1$ $$f_X,Y(x,y) = begincases2&:& 0<x<y<1\0&:&textsfotherwiseendcases$$

What do I do from there on?

Draw the line $x+y=1$ on your graph. Notice how it divides the triangle. The ratio of areas will give the probability.

$$mathsf P(X+Y<1) =dfraclvert Rcap(x,y)inBbb R^2:x+y<1rvertlvert R rvert$$

Or you might do it the long way. If you really want to:

$$beginalignmathsf P(X+Ylt 1)&=int_0^1int_0^min(y,1-y)2~mathrm d x~mathrm d y\[1ex]&=int_0^1/2int_0^y 2mathsf d xmathsf d y+int_1/2^1int_0^1-y 2mathsf d xmathsf d y\[1ex]&~~vdotsendalign$$

share|cite|improve this answer

answered Mar 27 at 3:42

Graham KempGraham Kemp

88k43579

$endgroup$

share|cite|improve this answer

answered Mar 27 at 3:42

Graham KempGraham Kemp

88k43579

answered Mar 27 at 3:42

Graham KempGraham Kemp

88k43579

answered Mar 27 at 3:42

Graham KempGraham Kemp

88k43579