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If $X$, $Y$ are i.i.d and mutually exclusive, then is $Pr(min(X,Y)=n)=Pr(X=n)+Pr(Y=n)$?

0

0

$begingroup$

I was wondering if $X$, $Y$ are independent and mutually exclusive, then is $Pr(min(X,Y)=n)=Pr(X=n)+Pr(Y=n)$?

$Pr(min(X,Y)=n)=Pr(X=n text{ or }Y=n)=Pr(X=n)+Pr(Y=n)$. Is this correct or am I missing something?

probability probability-theory

share|cite|improve this question

edited Mar 23 at 22:35

Rócherz

3,0263823

asked Mar 23 at 22:16

HeisenbergHeisenberg

1,3281742

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You might be confusing a random variable and an event
  $endgroup$
  – Vladislav
  Mar 23 at 22:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What I mean is X=n and Y=n are mutually exclusive events
  $endgroup$
  – Heisenberg
  Mar 23 at 22:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's not generally true that $$color{red}{Pr(min(X,Y)=n)=Pr(X=ntext{ or } Y=n)}.$$ You maybe be thinking that $min(X,Y)=n$ if and only if $X=n$ or $Y=n$, but this is not the case (if it were the case though, the above equation would be correct).
  $endgroup$
  – Minus One-Twelfth
  Mar 23 at 22:32
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then they cannot be iid.
  $endgroup$
  – kimchi lover
  Mar 23 at 22:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $min(X,Y)=n$ requires one of them $=n$ and the other $ge n$.
  $endgroup$
  – herb steinberg
  Mar 23 at 22:36
  

  
  
  
  

 | 
show 2 more comments

0

0

$begingroup$

I was wondering if $X$, $Y$ are independent and mutually exclusive, then is $Pr(min(X,Y)=n)=Pr(X=n)+Pr(Y=n)$?

$Pr(min(X,Y)=n)=Pr(X=n text{ or }Y=n)=Pr(X=n)+Pr(Y=n)$. Is this correct or am I missing something?

probability probability-theory

share|cite|improve this question

edited Mar 23 at 22:35

Rócherz

3,0263823

asked Mar 23 at 22:16

HeisenbergHeisenberg

1,3281742

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You might be confusing a random variable and an event
  $endgroup$
  – Vladislav
  Mar 23 at 22:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What I mean is X=n and Y=n are mutually exclusive events
  $endgroup$
  – Heisenberg
  Mar 23 at 22:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's not generally true that $$color{red}{Pr(min(X,Y)=n)=Pr(X=ntext{ or } Y=n)}.$$ You maybe be thinking that $min(X,Y)=n$ if and only if $X=n$ or $Y=n$, but this is not the case (if it were the case though, the above equation would be correct).
  $endgroup$
  – Minus One-Twelfth
  Mar 23 at 22:32
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then they cannot be iid.
  $endgroup$
  – kimchi lover
  Mar 23 at 22:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $min(X,Y)=n$ requires one of them $=n$ and the other $ge n$.
  $endgroup$
  – herb steinberg
  Mar 23 at 22:36
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

I was wondering if $X$, $Y$ are independent and mutually exclusive, then is $Pr(min(X,Y)=n)=Pr(X=n)+Pr(Y=n)$?

$Pr(min(X,Y)=n)=Pr(X=n text{ or }Y=n)=Pr(X=n)+Pr(Y=n)$. Is this correct or am I missing something?

probability probability-theory

share|cite|improve this question

edited Mar 23 at 22:35

Rócherz

3,0263823

asked Mar 23 at 22:16

HeisenbergHeisenberg

1,3281742

$endgroup$

share|cite|improve this question

edited Mar 23 at 22:35

Rócherz

3,0263823

asked Mar 23 at 22:16

HeisenbergHeisenberg

1,3281742

share|cite|improve this question

edited Mar 23 at 22:35

Rócherz

3,0263823

asked Mar 23 at 22:16

HeisenbergHeisenberg

1,3281742

edited Mar 23 at 22:35

Rócherz

3,0263823

edited Mar 23 at 22:35

Rócherz

3,0263823

asked Mar 23 at 22:16

HeisenbergHeisenberg

1,3281742

asked Mar 23 at 22:16

HeisenbergHeisenberg

1,3281742

2 Answers
2

active

oldest

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3

$begingroup$

$$Pr(min(X,Y)=n)=Pr(X=ncap Y=n)+Pr(X=ncap Ygt n)+Pr(Y=ncap Xgt n) ,$$ since these events are mutually exclusive.

Using independence $Pr(min(X,Y)=n)=Pr(X=n)Pr(Y=n)+Pr(X=n)Pr(Ygt n)+Pr(Y=n)Pr(Xgt n)$.

share|cite|improve this answer

edited Mar 24 at 3:38

lonza leggiera

1,457128

answered Mar 23 at 23:14

herb steinbergherb steinberg

3,1832311

$endgroup$

add a comment | 

2

$begingroup$

What you want to do is the following:
$$P(min(X,Y)geq n) = P(Xgeq n, Ygeq n) = P(Xgeq n)P(Ygeq n) = (1-F_X(n))(1-F_Y(n))$$
$$ = 1-F_X(n)-F_Y(n)+F_X(n)F_Y(n)$$
where $F_X$ is the cdf of $X.$
Then you have that,
$$P(min(X,Y) = n) = P(min(X,Y)geq n) - P(min(X,Y)geq n+1)$$
$$ = F_X(n+1)-F_X(n)+F_Y(n+1)-F_Y(n)+F_X(n)F_Y(n) - F_X(n+1)F_Y(n+1).$$

share|cite|improve this answer

answered Mar 23 at 22:46

model_checkermodel_checker

4,45521931

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add a comment | 

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3

$begingroup$

$$Pr(min(X,Y)=n)=Pr(X=ncap Y=n)+Pr(X=ncap Ygt n)+Pr(Y=ncap Xgt n) ,$$ since these events are mutually exclusive.

Using independence $Pr(min(X,Y)=n)=Pr(X=n)Pr(Y=n)+Pr(X=n)Pr(Ygt n)+Pr(Y=n)Pr(Xgt n)$.

share|cite|improve this answer

edited Mar 24 at 3:38

lonza leggiera

1,457128

answered Mar 23 at 23:14

herb steinbergherb steinberg

3,1832311

$endgroup$

add a comment | 

3

$begingroup$

$$Pr(min(X,Y)=n)=Pr(X=ncap Y=n)+Pr(X=ncap Ygt n)+Pr(Y=ncap Xgt n) ,$$ since these events are mutually exclusive.

Using independence $Pr(min(X,Y)=n)=Pr(X=n)Pr(Y=n)+Pr(X=n)Pr(Ygt n)+Pr(Y=n)Pr(Xgt n)$.

share|cite|improve this answer

edited Mar 24 at 3:38

lonza leggiera

1,457128

answered Mar 23 at 23:14

herb steinbergherb steinberg

3,1832311

$endgroup$

add a comment | 

$begingroup$

$$Pr(min(X,Y)=n)=Pr(X=ncap Y=n)+Pr(X=ncap Ygt n)+Pr(Y=ncap Xgt n) ,$$ since these events are mutually exclusive.

Using independence $Pr(min(X,Y)=n)=Pr(X=n)Pr(Y=n)+Pr(X=n)Pr(Ygt n)+Pr(Y=n)Pr(Xgt n)$.

share|cite|improve this answer

edited Mar 24 at 3:38

lonza leggiera

1,457128

answered Mar 23 at 23:14

herb steinbergherb steinberg

3,1832311

$endgroup$

share|cite|improve this answer

edited Mar 24 at 3:38

lonza leggiera

1,457128

answered Mar 23 at 23:14

herb steinbergherb steinberg

3,1832311

edited Mar 24 at 3:38

lonza leggiera

1,457128

edited Mar 24 at 3:38

lonza leggiera

1,457128

answered Mar 23 at 23:14

herb steinbergherb steinberg

3,1832311

answered Mar 23 at 23:14

herb steinbergherb steinberg

3,1832311

2

$begingroup$

What you want to do is the following:
$$P(min(X,Y)geq n) = P(Xgeq n, Ygeq n) = P(Xgeq n)P(Ygeq n) = (1-F_X(n))(1-F_Y(n))$$
$$ = 1-F_X(n)-F_Y(n)+F_X(n)F_Y(n)$$
where $F_X$ is the cdf of $X.$
Then you have that,
$$P(min(X,Y) = n) = P(min(X,Y)geq n) - P(min(X,Y)geq n+1)$$
$$ = F_X(n+1)-F_X(n)+F_Y(n+1)-F_Y(n)+F_X(n)F_Y(n) - F_X(n+1)F_Y(n+1).$$

share|cite|improve this answer

answered Mar 23 at 22:46

model_checkermodel_checker

4,45521931

$endgroup$

add a comment | 

2

$begingroup$

What you want to do is the following:
$$P(min(X,Y)geq n) = P(Xgeq n, Ygeq n) = P(Xgeq n)P(Ygeq n) = (1-F_X(n))(1-F_Y(n))$$
$$ = 1-F_X(n)-F_Y(n)+F_X(n)F_Y(n)$$
where $F_X$ is the cdf of $X.$
Then you have that,
$$P(min(X,Y) = n) = P(min(X,Y)geq n) - P(min(X,Y)geq n+1)$$
$$ = F_X(n+1)-F_X(n)+F_Y(n+1)-F_Y(n)+F_X(n)F_Y(n) - F_X(n+1)F_Y(n+1).$$

share|cite|improve this answer

answered Mar 23 at 22:46

model_checkermodel_checker

4,45521931

$endgroup$

add a comment | 

$begingroup$

What you want to do is the following:
$$P(min(X,Y)geq n) = P(Xgeq n, Ygeq n) = P(Xgeq n)P(Ygeq n) = (1-F_X(n))(1-F_Y(n))$$
$$ = 1-F_X(n)-F_Y(n)+F_X(n)F_Y(n)$$
where $F_X$ is the cdf of $X.$
Then you have that,
$$P(min(X,Y) = n) = P(min(X,Y)geq n) - P(min(X,Y)geq n+1)$$
$$ = F_X(n+1)-F_X(n)+F_Y(n+1)-F_Y(n)+F_X(n)F_Y(n) - F_X(n+1)F_Y(n+1).$$

share|cite|improve this answer

answered Mar 23 at 22:46

model_checkermodel_checker

4,45521931

$endgroup$

share|cite|improve this answer

answered Mar 23 at 22:46

model_checkermodel_checker

4,45521931

answered Mar 23 at 22:46

model_checkermodel_checker

4,45521931

answered Mar 23 at 22:46

model_checkermodel_checker

4,45521931