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2

$begingroup$

Let $(X_i)_i=1,dots,n$ be a finite sequence of random variables such that $X_isimmathcalE(lambda_i).$

We can prove that $Y:=min_1le ile nX_isimmathcalE(lambda=sum_i=1^n lambda_i)$

Now I would like to compute $P(X_i=Y).$

We have
$$P(min_jne iX_j>x)=P(cap_jne iX_j>x)=prod_jne je^-lambda_j x=e^-(lambda-lambda_i)x$$

Now not sure how I can continue.

probability-distributions random-variables exponential-distribution

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asked Mar 15 at 15:59

JulienJulien

1205

$endgroup$

add a comment | 

2

$begingroup$

Let $(X_i)_i=1,dots,n$ be a finite sequence of random variables such that $X_isimmathcalE(lambda_i).$

We can prove that $Y:=min_1le ile nX_isimmathcalE(lambda=sum_i=1^n lambda_i)$

Now I would like to compute $P(X_i=Y).$

We have
$$P(min_jne iX_j>x)=P(cap_jne iX_j>x)=prod_jne je^-lambda_j x=e^-(lambda-lambda_i)x$$

Now not sure how I can continue.

probability-distributions random-variables exponential-distribution

share|cite|improve this question

asked Mar 15 at 15:59

JulienJulien

1205

$endgroup$

add a comment | 

$begingroup$

Let $(X_i)_i=1,dots,n$ be a finite sequence of random variables such that $X_isimmathcalE(lambda_i).$

We can prove that $Y:=min_1le ile nX_isimmathcalE(lambda=sum_i=1^n lambda_i)$

Now I would like to compute $P(X_i=Y).$

We have
$$P(min_jne iX_j>x)=P(cap_jne iX_j>x)=prod_jne je^-lambda_j x=e^-(lambda-lambda_i)x$$

Now not sure how I can continue.

probability-distributions random-variables exponential-distribution

share|cite|improve this question

asked Mar 15 at 15:59

JulienJulien

1205

$endgroup$

share|cite|improve this question

asked Mar 15 at 15:59

JulienJulien

1205

share|cite|improve this question

asked Mar 15 at 15:59

JulienJulien

1205

asked Mar 15 at 15:59

JulienJulien

1205

asked Mar 15 at 15:59

JulienJulien

1205

1 Answer
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0

$begingroup$

Manipulating the probabilities:

$$P(X_i = Y) = 1 - P(X_i ne Y) = 1 - (P(X_i < Y) + P(X_i > Y)) = 1 - P(X_i > Y) =$$

$$1 - P(X_i > textmin_j = 1, ldots, nX_j) = 1 - P(X_i > textmin_j ne iX_j)$$

Now, $X_i$ and $Y_i = textmin_j ne iX_j$ are two independent exponential random variables with parameters $lambda_i$ and $tildelambda_i = sum_i ne j lambda_j$. Then $P(X_i > Y_i) = P(mathcalE(lambda_i) > mathcalE(tildelambda_i)) = tildelambda_i / (lambda_i + tildelambda_i) = tildelambda_i / lambda$.

So $P(X_i = Y) = 1 - tildelambda_i / lambda = (lambda - tildelambda_i) / lambda = lambda_i / lambda$.

share|cite|improve this answer

edited Mar 15 at 17:15

answered Mar 15 at 17:07

dcolazindcolazin

3694

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks but I don't understand what do you mean by $P(mathcalE(lambda_i) > mathcalE(tildelambda_i)) $?
  $endgroup$
  – Julien
  Mar 17 at 7:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Julien it was just a fast way to denote $P(Z>W)$, where $Z = mathcalE(z), W = mathcalE(w)$ and are independent (you can find the calculation here)
  $endgroup$
  – dcolazin
  Mar 17 at 7:56
  

  
  
  
  

add a comment | 

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0

$begingroup$

Manipulating the probabilities:

$$P(X_i = Y) = 1 - P(X_i ne Y) = 1 - (P(X_i < Y) + P(X_i > Y)) = 1 - P(X_i > Y) =$$

$$1 - P(X_i > textmin_j = 1, ldots, nX_j) = 1 - P(X_i > textmin_j ne iX_j)$$

Now, $X_i$ and $Y_i = textmin_j ne iX_j$ are two independent exponential random variables with parameters $lambda_i$ and $tildelambda_i = sum_i ne j lambda_j$. Then $P(X_i > Y_i) = P(mathcalE(lambda_i) > mathcalE(tildelambda_i)) = tildelambda_i / (lambda_i + tildelambda_i) = tildelambda_i / lambda$.

So $P(X_i = Y) = 1 - tildelambda_i / lambda = (lambda - tildelambda_i) / lambda = lambda_i / lambda$.

share|cite|improve this answer

edited Mar 15 at 17:15

answered Mar 15 at 17:07

dcolazindcolazin

3694

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks but I don't understand what do you mean by $P(mathcalE(lambda_i) > mathcalE(tildelambda_i)) $?
  $endgroup$
  – Julien
  Mar 17 at 7:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Julien it was just a fast way to denote $P(Z>W)$, where $Z = mathcalE(z), W = mathcalE(w)$ and are independent (you can find the calculation here)
  $endgroup$
  – dcolazin
  Mar 17 at 7:56
  

  
  
  
  

add a comment | 

0

$begingroup$

Manipulating the probabilities:

$$P(X_i = Y) = 1 - P(X_i ne Y) = 1 - (P(X_i < Y) + P(X_i > Y)) = 1 - P(X_i > Y) =$$

$$1 - P(X_i > textmin_j = 1, ldots, nX_j) = 1 - P(X_i > textmin_j ne iX_j)$$

Now, $X_i$ and $Y_i = textmin_j ne iX_j$ are two independent exponential random variables with parameters $lambda_i$ and $tildelambda_i = sum_i ne j lambda_j$. Then $P(X_i > Y_i) = P(mathcalE(lambda_i) > mathcalE(tildelambda_i)) = tildelambda_i / (lambda_i + tildelambda_i) = tildelambda_i / lambda$.

So $P(X_i = Y) = 1 - tildelambda_i / lambda = (lambda - tildelambda_i) / lambda = lambda_i / lambda$.

share|cite|improve this answer

edited Mar 15 at 17:15

answered Mar 15 at 17:07

dcolazindcolazin

3694

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks but I don't understand what do you mean by $P(mathcalE(lambda_i) > mathcalE(tildelambda_i)) $?
  $endgroup$
  – Julien
  Mar 17 at 7:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Julien it was just a fast way to denote $P(Z>W)$, where $Z = mathcalE(z), W = mathcalE(w)$ and are independent (you can find the calculation here)
  $endgroup$
  – dcolazin
  Mar 17 at 7:56
  

  
  
  
  

add a comment | 

$begingroup$

Manipulating the probabilities:

$$P(X_i = Y) = 1 - P(X_i ne Y) = 1 - (P(X_i < Y) + P(X_i > Y)) = 1 - P(X_i > Y) =$$

$$1 - P(X_i > textmin_j = 1, ldots, nX_j) = 1 - P(X_i > textmin_j ne iX_j)$$

Now, $X_i$ and $Y_i = textmin_j ne iX_j$ are two independent exponential random variables with parameters $lambda_i$ and $tildelambda_i = sum_i ne j lambda_j$. Then $P(X_i > Y_i) = P(mathcalE(lambda_i) > mathcalE(tildelambda_i)) = tildelambda_i / (lambda_i + tildelambda_i) = tildelambda_i / lambda$.

So $P(X_i = Y) = 1 - tildelambda_i / lambda = (lambda - tildelambda_i) / lambda = lambda_i / lambda$.

share|cite|improve this answer

edited Mar 15 at 17:15

answered Mar 15 at 17:07

dcolazindcolazin

3694

$endgroup$

share|cite|improve this answer

edited Mar 15 at 17:15

answered Mar 15 at 17:07

dcolazindcolazin

3694

answered Mar 15 at 17:07

dcolazindcolazin

3694

answered Mar 15 at 17:07

dcolazindcolazin

3694