Find expected value for $max (X_1,…,X_n)$ [duplicate]

0

$begingroup$

This question already has an answer here:

• 
  Expected value of $max{X_1,ldots,X_n}$ where $X_i$ are iid uniform.
  
  2 answers
  
  

I have a random variable $X$ behave as $ U[0,alpha]$.

Out of $n$ experiments, I would look at the new random variable $Y=max(X_1,...X_n)$.

I ask for $E(Y)$.

I know it supposed to be $alpha cdot frac{n-1}{n}$ or something like that (maybe $alpha cdot frac {n}{n+1}$) and I understand why, but all this is intuitive and I don't understand how to prove it.

Thanks in advance for your answers

probability probability-distributions expected-value

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edited Jan 30 at 14:05

TPace

5111318

asked Jan 30 at 13:55

ShaqShaq

3049

$endgroup$

marked as duplicate by Community♦ Jan 30 at 14:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are the $X_i$'s independent ?
  $endgroup$
  – Surb
  Jan 30 at 13:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  just Google it jamesmccammon.com/2017/02/18/…
  $endgroup$
  – Paul
  Jan 30 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes every $X_i$ is independent and is U[0,$alpha$]
  $endgroup$
  – Shaq
  Jan 30 at 14:00
  

  
  
  
  

add a comment | 

0

$begingroup$

This question already has an answer here:

• 
  Expected value of $max{X_1,ldots,X_n}$ where $X_i$ are iid uniform.
  
  2 answers
  
  

I have a random variable $X$ behave as $ U[0,alpha]$.

Out of $n$ experiments, I would look at the new random variable $Y=max(X_1,...X_n)$.

I ask for $E(Y)$.

I know it supposed to be $alpha cdot frac{n-1}{n}$ or something like that (maybe $alpha cdot frac {n}{n+1}$) and I understand why, but all this is intuitive and I don't understand how to prove it.

Thanks in advance for your answers

probability probability-distributions expected-value

share|cite|improve this question

edited Jan 30 at 14:05

TPace

5111318

asked Jan 30 at 13:55

ShaqShaq

3049

$endgroup$

marked as duplicate by Community♦ Jan 30 at 14:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are the $X_i$'s independent ?
  $endgroup$
  – Surb
  Jan 30 at 13:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  just Google it jamesmccammon.com/2017/02/18/…
  $endgroup$
  – Paul
  Jan 30 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes every $X_i$ is independent and is U[0,$alpha$]
  $endgroup$
  – Shaq
  Jan 30 at 14:00
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

This question already has an answer here:

• 
  Expected value of $max{X_1,ldots,X_n}$ where $X_i$ are iid uniform.
  
  2 answers
  
  

I have a random variable $X$ behave as $ U[0,alpha]$.

Out of $n$ experiments, I would look at the new random variable $Y=max(X_1,...X_n)$.

I ask for $E(Y)$.

I know it supposed to be $alpha cdot frac{n-1}{n}$ or something like that (maybe $alpha cdot frac {n}{n+1}$) and I understand why, but all this is intuitive and I don't understand how to prove it.

Thanks in advance for your answers

probability probability-distributions expected-value

share|cite|improve this question

edited Jan 30 at 14:05

TPace

5111318

asked Jan 30 at 13:55

ShaqShaq

3049

$endgroup$

This question already has an answer here:

• 
  Expected value of $max{X_1,ldots,X_n}$ where $X_i$ are iid uniform.
  
  2 answers
  
  

I have a random variable $X$ behave as $ U[0,alpha]$.

Out of $n$ experiments, I would look at the new random variable $Y=max(X_1,...X_n)$.

I ask for $E(Y)$.

I know it supposed to be $alpha cdot frac{n-1}{n}$ or something like that (maybe $alpha cdot frac {n}{n+1}$) and I understand why, but all this is intuitive and I don't understand how to prove it.

Thanks in advance for your answers

This question already has an answer here:

• 
  Expected value of $max{X_1,ldots,X_n}$ where $X_i$ are iid uniform.
  
  2 answers
  
  

probability probability-distributions expected-value

probability probability-distributions expected-value

share|cite|improve this question

edited Jan 30 at 14:05

TPace

5111318

asked Jan 30 at 13:55

ShaqShaq

3049

share|cite|improve this question

edited Jan 30 at 14:05

TPace

5111318

asked Jan 30 at 13:55

ShaqShaq

3049

share|cite|improve this question

share|cite|improve this question

edited Jan 30 at 14:05

TPace

5111318

edited Jan 30 at 14:05

TPace

5111318

edited Jan 30 at 14:05

TPace

5111318

5111318

asked Jan 30 at 13:55

ShaqShaq

3049

asked Jan 30 at 13:55

ShaqShaq

3049

asked Jan 30 at 13:55

ShaqShaq

3049

3049

marked as duplicate by Community♦ Jan 30 at 14:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Community♦ Jan 30 at 14:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are the $X_i$'s independent ?
  $endgroup$
  – Surb
  Jan 30 at 13:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  just Google it jamesmccammon.com/2017/02/18/…
  $endgroup$
  – Paul
  Jan 30 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes every $X_i$ is independent and is U[0,$alpha$]
  $endgroup$
  – Shaq
  Jan 30 at 14:00
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are the $X_i$'s independent ?
  $endgroup$
  – Surb
  Jan 30 at 13:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  just Google it jamesmccammon.com/2017/02/18/…
  $endgroup$
  – Paul
  Jan 30 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes every $X_i$ is independent and is U[0,$alpha$]
  $endgroup$
  – Shaq
  Jan 30 at 14:00
  

  
  
  
  

$begingroup$
Are the $X_i$'s independent ?
$endgroup$
– Surb
Jan 30 at 13:59

$begingroup$
Are the $X_i$'s independent ?
$endgroup$
– Surb
Jan 30 at 13:59

1

1

$begingroup$
just Google it jamesmccammon.com/2017/02/18/…
$endgroup$
– Paul
Jan 30 at 13:59

$begingroup$
just Google it jamesmccammon.com/2017/02/18/…
$endgroup$
– Paul
Jan 30 at 13:59

$begingroup$
Yes every $X_i$ is independent and is U[0,$alpha$]
$endgroup$
– Shaq
Jan 30 at 14:00

$begingroup$
Yes every $X_i$ is independent and is U[0,$alpha$]
$endgroup$
– Shaq
Jan 30 at 14:00

add a comment | 

2 Answers
2

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2

$begingroup$

Let $$Z = max(X_1,X_2,...X_n)$$

$$P(Zle z) = P(max(X_1,X_2,...X_n)le z)$$

$$= F(Xle z)^{n}$$

$$ = (frac{z}{alpha})^{n}$$

pdf is the derivative of it

Hence $$f_Z(z) = frac{n.z^{n-1}}{alpha^n}$$

$$E(Z) = int_{0}^{alpha} frac{nz..z^{n-1}}{alpha^n} = frac{nalpha}{(n+1)}$$

share|cite|improve this answer

edited Jan 30 at 14:12

answered Jan 30 at 14:07

Satish RamanathanSatish Ramanathan

10k31323

$endgroup$

add a comment | 

3

$begingroup$

$$mathbb P{Yleq y}=mathbb P{X_1leq y,...,X_nleq y}=mathbb P{X_1leq y}^n=F_{X_1}(y)^n.$$
Therefore $$f_Y(y)=nf_{X_1}(y)F_{X_1}(y)^{n-1}=nfrac{1}{alpha^n }y^{n-1}boldsymbol 1_{[0,alpha ]}(y).$$

Finally, $$mathbb E[Y]=int_0^alpha yf_Y(y),mathrm d x=...$$

share|cite|improve this answer

edited Jan 30 at 14:10

answered Jan 30 at 14:04

SurbSurb

38.4k94478

$endgroup$

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

$begingroup$

Let $$Z = max(X_1,X_2,...X_n)$$

$$P(Zle z) = P(max(X_1,X_2,...X_n)le z)$$

$$= F(Xle z)^{n}$$

$$ = (frac{z}{alpha})^{n}$$

pdf is the derivative of it

Hence $$f_Z(z) = frac{n.z^{n-1}}{alpha^n}$$

$$E(Z) = int_{0}^{alpha} frac{nz..z^{n-1}}{alpha^n} = frac{nalpha}{(n+1)}$$

share|cite|improve this answer

edited Jan 30 at 14:12

answered Jan 30 at 14:07

Satish RamanathanSatish Ramanathan

10k31323

$endgroup$

add a comment | 

2

$begingroup$

Let $$Z = max(X_1,X_2,...X_n)$$

$$P(Zle z) = P(max(X_1,X_2,...X_n)le z)$$

$$= F(Xle z)^{n}$$

$$ = (frac{z}{alpha})^{n}$$

pdf is the derivative of it

Hence $$f_Z(z) = frac{n.z^{n-1}}{alpha^n}$$

$$E(Z) = int_{0}^{alpha} frac{nz..z^{n-1}}{alpha^n} = frac{nalpha}{(n+1)}$$

share|cite|improve this answer

edited Jan 30 at 14:12

answered Jan 30 at 14:07

Satish RamanathanSatish Ramanathan

10k31323

$endgroup$

add a comment | 

2

2

2

$begingroup$

Let $$Z = max(X_1,X_2,...X_n)$$

$$P(Zle z) = P(max(X_1,X_2,...X_n)le z)$$

$$= F(Xle z)^{n}$$

$$ = (frac{z}{alpha})^{n}$$

pdf is the derivative of it

Hence $$f_Z(z) = frac{n.z^{n-1}}{alpha^n}$$

$$E(Z) = int_{0}^{alpha} frac{nz..z^{n-1}}{alpha^n} = frac{nalpha}{(n+1)}$$

share|cite|improve this answer

edited Jan 30 at 14:12

answered Jan 30 at 14:07

Satish RamanathanSatish Ramanathan

10k31323

$endgroup$

Let $$Z = max(X_1,X_2,...X_n)$$

$$P(Zle z) = P(max(X_1,X_2,...X_n)le z)$$

$$= F(Xle z)^{n}$$

$$ = (frac{z}{alpha})^{n}$$

pdf is the derivative of it

Hence $$f_Z(z) = frac{n.z^{n-1}}{alpha^n}$$

$$E(Z) = int_{0}^{alpha} frac{nz..z^{n-1}}{alpha^n} = frac{nalpha}{(n+1)}$$

share|cite|improve this answer

edited Jan 30 at 14:12

answered Jan 30 at 14:07

Satish RamanathanSatish Ramanathan

10k31323

share|cite|improve this answer

share|cite|improve this answer

edited Jan 30 at 14:12

edited Jan 30 at 14:12

edited Jan 30 at 14:12

answered Jan 30 at 14:07

Satish RamanathanSatish Ramanathan

10k31323

answered Jan 30 at 14:07

Satish RamanathanSatish Ramanathan

10k31323

answered Jan 30 at 14:07

Satish RamanathanSatish Ramanathan

10k31323

10k31323

add a comment | 

add a comment | 

3

$begingroup$

$$mathbb P{Yleq y}=mathbb P{X_1leq y,...,X_nleq y}=mathbb P{X_1leq y}^n=F_{X_1}(y)^n.$$
Therefore $$f_Y(y)=nf_{X_1}(y)F_{X_1}(y)^{n-1}=nfrac{1}{alpha^n }y^{n-1}boldsymbol 1_{[0,alpha ]}(y).$$

Finally, $$mathbb E[Y]=int_0^alpha yf_Y(y),mathrm d x=...$$

share|cite|improve this answer

edited Jan 30 at 14:10

answered Jan 30 at 14:04

SurbSurb

38.4k94478

$endgroup$

add a comment | 

3

$begingroup$

$$mathbb P{Yleq y}=mathbb P{X_1leq y,...,X_nleq y}=mathbb P{X_1leq y}^n=F_{X_1}(y)^n.$$
Therefore $$f_Y(y)=nf_{X_1}(y)F_{X_1}(y)^{n-1}=nfrac{1}{alpha^n }y^{n-1}boldsymbol 1_{[0,alpha ]}(y).$$

Finally, $$mathbb E[Y]=int_0^alpha yf_Y(y),mathrm d x=...$$

share|cite|improve this answer

edited Jan 30 at 14:10

answered Jan 30 at 14:04

SurbSurb

38.4k94478

$endgroup$

add a comment | 

3

3

3

$begingroup$

$$mathbb P{Yleq y}=mathbb P{X_1leq y,...,X_nleq y}=mathbb P{X_1leq y}^n=F_{X_1}(y)^n.$$
Therefore $$f_Y(y)=nf_{X_1}(y)F_{X_1}(y)^{n-1}=nfrac{1}{alpha^n }y^{n-1}boldsymbol 1_{[0,alpha ]}(y).$$

Finally, $$mathbb E[Y]=int_0^alpha yf_Y(y),mathrm d x=...$$

share|cite|improve this answer

edited Jan 30 at 14:10

answered Jan 30 at 14:04

SurbSurb

38.4k94478

$endgroup$

$$mathbb P{Yleq y}=mathbb P{X_1leq y,...,X_nleq y}=mathbb P{X_1leq y}^n=F_{X_1}(y)^n.$$
Therefore $$f_Y(y)=nf_{X_1}(y)F_{X_1}(y)^{n-1}=nfrac{1}{alpha^n }y^{n-1}boldsymbol 1_{[0,alpha ]}(y).$$

Finally, $$mathbb E[Y]=int_0^alpha yf_Y(y),mathrm d x=...$$

share|cite|improve this answer

edited Jan 30 at 14:10

answered Jan 30 at 14:04

SurbSurb

38.4k94478

share|cite|improve this answer

share|cite|improve this answer

edited Jan 30 at 14:10

edited Jan 30 at 14:10

edited Jan 30 at 14:10

answered Jan 30 at 14:04

SurbSurb

38.4k94478

answered Jan 30 at 14:04

SurbSurb

38.4k94478

answered Jan 30 at 14:04

SurbSurb

38.4k94478

38.4k94478

add a comment | 

add a comment | 

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