Expected value of a prize that has a geometric distribution dependent on independent draws from a uniform...
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A number $X$ and a sequence of numbers ${Y_nvert nin mathbb N}$ are i.i.d draws from the uniform distribution on $[0,1]$. Let $N = inf{n in mathbb Nvert Y_n > X}$. The player conducting these draws receives the prize of $(N - 1)$. Calculate the expected value of this prize.
I understand that $(N-1)/X= x sim Geometric(1 - X)$ but I do not know how to proceed further.
statistics probability-distributions conditional-expectation
asked Dec 6 '18 at 14:18
Chetna AhujaChetna Ahuja
12
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add a comment |
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A number $X$ and a sequence of numbers ${Y_nvert nin mathbb N}$ are i.i.d draws from the uniform distribution on $[0,1]$. Let $N = inf{n in mathbb Nvert Y_n > X}$. The player conducting these draws receives the prize of $(N - 1)$. Calculate the expected value of this prize.
I understand that $(N-1)/X= x sim Geometric(1 - X)$ but I do not know how to proceed further.
statistics probability-distributions conditional-expectation
asked Dec 6 '18 at 14:18
Chetna AhujaChetna Ahuja
12
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1
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Welcome to Math.SE! Please update your question to include your thoughts on the problem, and where you are getting stuck exactly. We will be happy to guide you further.
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– gt6989b
Dec 6 '18 at 14:24
add a comment |
$begingroup$
A number $X$ and a sequence of numbers ${Y_nvert nin mathbb N}$ are i.i.d draws from the uniform distribution on $[0,1]$. Let $N = inf{n in mathbb Nvert Y_n > X}$. The player conducting these draws receives the prize of $(N - 1)$. Calculate the expected value of this prize.
I understand that $(N-1)/X= x sim Geometric(1 - X)$ but I do not know how to proceed further.
statistics probability-distributions conditional-expectation
asked Dec 6 '18 at 14:18
Chetna AhujaChetna Ahuja
12
$endgroup$
A number $X$ and a sequence of numbers ${Y_nvert nin mathbb N}$ are i.i.d draws from the uniform distribution on $[0,1]$. Let $N = inf{n in mathbb Nvert Y_n > X}$. The player conducting these draws receives the prize of $(N - 1)$. Calculate the expected value of this prize.
I understand that $(N-1)/X= x sim Geometric(1 - X)$ but I do not know how to proceed further.
statistics probability-distributions conditional-expectation
statistics probability-distributions conditional-expectation
asked Dec 6 '18 at 14:18
Chetna AhujaChetna Ahuja
12
asked Dec 6 '18 at 14:18
Chetna AhujaChetna Ahuja
12
edited Dec 6 '18 at 14:39
Chetna Ahuja
asked Dec 6 '18 at 14:18
Chetna AhujaChetna Ahuja
12
asked Dec 6 '18 at 14:18
Chetna AhujaChetna Ahuja
12
asked Dec 6 '18 at 14:18
Chetna AhujaChetna Ahuja
12
12
1
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Welcome to Math.SE! Please update your question to include your thoughts on the problem, and where you are getting stuck exactly. We will be happy to guide you further.
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– gt6989b
Dec 6 '18 at 14:24
add a comment |
1
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Welcome to Math.SE! Please update your question to include your thoughts on the problem, and where you are getting stuck exactly. We will be happy to guide you further.
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– gt6989b
Dec 6 '18 at 14:24
1
1
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Welcome to Math.SE! Please update your question to include your thoughts on the problem, and where you are getting stuck exactly. We will be happy to guide you further.
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– gt6989b
Dec 6 '18 at 14:24
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Welcome to Math.SE! Please update your question to include your thoughts on the problem, and where you are getting stuck exactly. We will be happy to guide you further.
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– gt6989b
Dec 6 '18 at 14:24
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1 Answer
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Hint:
Observe that: $${N=n}={Y_n>Xgeqmax(Y_1,dots,Y_{n-1})}$$
concerning the order of $n+1$ iid and continuous random variables.
The orders for $X,Y_1,dots,Y_n$ are equiprobable and there are $(n+1)!$ orders.
Further there are $(n-1)!$ orders that satisfy the condition $Y_n>X>max(Y_1,dots,Y_{n-1})$ leading to: $$P(N=n)=frac{(n-1)!}{(n+1)!}=frac1{n(n+1)}$$
answered Dec 6 '18 at 15:45
drhabdrhab
100k544130
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1
$begingroup$
Even simpler: ${N>n}={X>max{Y_1,ldots,Y_n}}$. By the exchangeability of the sample $(Y_1,ldots,Y_n,X)$, the probability of this event is $frac1{n+1}$, end of proof.
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– Did
Dec 6 '18 at 16:27
add a comment |
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1 Answer
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$begingroup$
Hint:
Observe that: $${N=n}={Y_n>Xgeqmax(Y_1,dots,Y_{n-1})}$$
concerning the order of $n+1$ iid and continuous random variables.
The orders for $X,Y_1,dots,Y_n$ are equiprobable and there are $(n+1)!$ orders.
Further there are $(n-1)!$ orders that satisfy the condition $Y_n>X>max(Y_1,dots,Y_{n-1})$ leading to: $$P(N=n)=frac{(n-1)!}{(n+1)!}=frac1{n(n+1)}$$
answered Dec 6 '18 at 15:45
drhabdrhab
100k544130
$endgroup$
1
$begingroup$
Even simpler: ${N>n}={X>max{Y_1,ldots,Y_n}}$. By the exchangeability of the sample $(Y_1,ldots,Y_n,X)$, the probability of this event is $frac1{n+1}$, end of proof.
$endgroup$
– Did
Dec 6 '18 at 16:27
add a comment |
$begingroup$
Hint:
Observe that: $${N=n}={Y_n>Xgeqmax(Y_1,dots,Y_{n-1})}$$
concerning the order of $n+1$ iid and continuous random variables.
The orders for $X,Y_1,dots,Y_n$ are equiprobable and there are $(n+1)!$ orders.
Further there are $(n-1)!$ orders that satisfy the condition $Y_n>X>max(Y_1,dots,Y_{n-1})$ leading to: $$P(N=n)=frac{(n-1)!}{(n+1)!}=frac1{n(n+1)}$$
answered Dec 6 '18 at 15:45
drhabdrhab
100k544130
$endgroup$
1
$begingroup$
Even simpler: ${N>n}={X>max{Y_1,ldots,Y_n}}$. By the exchangeability of the sample $(Y_1,ldots,Y_n,X)$, the probability of this event is $frac1{n+1}$, end of proof.
$endgroup$
– Did
Dec 6 '18 at 16:27
add a comment |
$begingroup$
Hint:
Observe that: $${N=n}={Y_n>Xgeqmax(Y_1,dots,Y_{n-1})}$$
concerning the order of $n+1$ iid and continuous random variables.
The orders for $X,Y_1,dots,Y_n$ are equiprobable and there are $(n+1)!$ orders.
Further there are $(n-1)!$ orders that satisfy the condition $Y_n>X>max(Y_1,dots,Y_{n-1})$ leading to: $$P(N=n)=frac{(n-1)!}{(n+1)!}=frac1{n(n+1)}$$
answered Dec 6 '18 at 15:45
drhabdrhab
100k544130
$endgroup$
Hint:
Observe that: $${N=n}={Y_n>Xgeqmax(Y_1,dots,Y_{n-1})}$$
concerning the order of $n+1$ iid and continuous random variables.
The orders for $X,Y_1,dots,Y_n$ are equiprobable and there are $(n+1)!$ orders.
Further there are $(n-1)!$ orders that satisfy the condition $Y_n>X>max(Y_1,dots,Y_{n-1})$ leading to: $$P(N=n)=frac{(n-1)!}{(n+1)!}=frac1{n(n+1)}$$
answered Dec 6 '18 at 15:45
drhabdrhab
100k544130
edited Dec 6 '18 at 16:52
answered Dec 6 '18 at 15:45
drhabdrhab
100k544130
answered Dec 6 '18 at 15:45
drhabdrhab
100k544130
answered Dec 6 '18 at 15:45
drhabdrhab
100k544130
100k544130
1
$begingroup$
Even simpler: ${N>n}={X>max{Y_1,ldots,Y_n}}$. By the exchangeability of the sample $(Y_1,ldots,Y_n,X)$, the probability of this event is $frac1{n+1}$, end of proof.
$endgroup$
– Did
Dec 6 '18 at 16:27
add a comment |
1
$begingroup$
Even simpler: ${N>n}={X>max{Y_1,ldots,Y_n}}$. By the exchangeability of the sample $(Y_1,ldots,Y_n,X)$, the probability of this event is $frac1{n+1}$, end of proof.
$endgroup$
– Did
Dec 6 '18 at 16:27
1
1
$begingroup$
Even simpler: ${N>n}={X>max{Y_1,ldots,Y_n}}$. By the exchangeability of the sample $(Y_1,ldots,Y_n,X)$, the probability of this event is $frac1{n+1}$, end of proof.
$endgroup$
– Did
Dec 6 '18 at 16:27
$begingroup$
Even simpler: ${N>n}={X>max{Y_1,ldots,Y_n}}$. By the exchangeability of the sample $(Y_1,ldots,Y_n,X)$, the probability of this event is $frac1{n+1}$, end of proof.
$endgroup$
– Did
Dec 6 '18 at 16:27
add a comment |
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Welcome to Math.SE! Please update your question to include your thoughts on the problem, and where you are getting stuck exactly. We will be happy to guide you further.
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– gt6989b
Dec 6 '18 at 14:24