Expected value of the shifted inverse of a binomial random variable, and applicationRandom variable and probability space notionsExpected value of a sum starting at a value given through a random variableExpected value complex random variableExpected value of a series of random variablesWhat to do if a random variable is apparently binomial but the amount of calculations required is too high?Continued Fraction and Random VariableProbability of correct answer at multiple choice testExpected value random variableExpected value of time between successes for binomial random variableFinding expected value of a stopping time dependent on a Poisson process and a variable $n$
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Expected value of the shifted inverse of a binomial random variable, and application
Random variable and probability space notionsExpected value of a sum starting at a value given through a random variableExpected value complex random variableExpected value of a series of random variablesWhat to do if a random variable is apparently binomial but the amount of calculations required is too high?Continued Fraction and Random VariableProbability of correct answer at multiple choice testExpected value random variableExpected value of time between successes for binomial random variableFinding expected value of a stopping time dependent on a Poisson process and a variable $n$
$begingroup$
Here is an exercise given by a colleague to a student :
Let $Xhookrightarrow B(n,p)$ and $Y=frac1X+1$. Find $rm E(Y)$.
It is not very difficult to prove that the answer is
$$rm E(Y) = frac1-q^n+1p(n+1)$$
where $q=1-p$. But the answer can also be written
$$rm E(Y) = frac1+q+q^2+dots+q^nn+1$$
First question: Is there any meaning to this form, which looks very much like a mean value of some sort? Or maybe another proof of this result which explains it in a more direct way?
Second question : Is there some context which could make this exercise more "concrete"?
probability-theory random-variables uniform-distribution binomial-distribution
edited Jun 21 '18 at 5:58
Did
249k23227466
asked Jun 21 '18 at 2:16
Nicolas FRANCOISNicolas FRANCOIS
3,7771516
$endgroup$
add a comment |
$begingroup$
Here is an exercise given by a colleague to a student :
Let $Xhookrightarrow B(n,p)$ and $Y=frac1X+1$. Find $rm E(Y)$.
It is not very difficult to prove that the answer is
$$rm E(Y) = frac1-q^n+1p(n+1)$$
where $q=1-p$. But the answer can also be written
$$rm E(Y) = frac1+q+q^2+dots+q^nn+1$$
First question: Is there any meaning to this form, which looks very much like a mean value of some sort? Or maybe another proof of this result which explains it in a more direct way?
Second question : Is there some context which could make this exercise more "concrete"?
probability-theory random-variables uniform-distribution binomial-distribution
edited Jun 21 '18 at 5:58
Did
249k23227466
asked Jun 21 '18 at 2:16
Nicolas FRANCOISNicolas FRANCOIS
3,7771516
$endgroup$
3
$begingroup$
In other words, is there a "natural" (whatever that means) explanation of the identity $$Eleft(frac1X+1right)=E(q^U)$$ for $X$ binomial $(n,p)$, $U$ uniform on $0,1,ldots,n$, and $q=1-p$? Interesting question...
$endgroup$
– Did
Jun 21 '18 at 5:55
$begingroup$
One can possibly justify the interchange of integral and summation to say that $Eleft(frac11+Xright)=Eleft(int_0^1 t^X,dtright)=int_0^1 E(t^X),dt=int_0^1 (1-p+pt)^n,dt$. But that's just another proof, not sure if this might be helpful in this regard.
$endgroup$
– StubbornAtom
Jun 21 '18 at 11:48
add a comment |
$begingroup$
Here is an exercise given by a colleague to a student :
Let $Xhookrightarrow B(n,p)$ and $Y=frac1X+1$. Find $rm E(Y)$.
It is not very difficult to prove that the answer is
$$rm E(Y) = frac1-q^n+1p(n+1)$$
where $q=1-p$. But the answer can also be written
$$rm E(Y) = frac1+q+q^2+dots+q^nn+1$$
First question: Is there any meaning to this form, which looks very much like a mean value of some sort? Or maybe another proof of this result which explains it in a more direct way?
Second question : Is there some context which could make this exercise more "concrete"?
probability-theory random-variables uniform-distribution binomial-distribution
edited Jun 21 '18 at 5:58
Did
249k23227466
asked Jun 21 '18 at 2:16
Nicolas FRANCOISNicolas FRANCOIS
3,7771516
$endgroup$
Here is an exercise given by a colleague to a student :
Let $Xhookrightarrow B(n,p)$ and $Y=frac1X+1$. Find $rm E(Y)$.
It is not very difficult to prove that the answer is
$$rm E(Y) = frac1-q^n+1p(n+1)$$
where $q=1-p$. But the answer can also be written
$$rm E(Y) = frac1+q+q^2+dots+q^nn+1$$
First question: Is there any meaning to this form, which looks very much like a mean value of some sort? Or maybe another proof of this result which explains it in a more direct way?
Second question : Is there some context which could make this exercise more "concrete"?
probability-theory random-variables uniform-distribution binomial-distribution
probability-theory random-variables uniform-distribution binomial-distribution
edited Jun 21 '18 at 5:58
Did
249k23227466
asked Jun 21 '18 at 2:16
Nicolas FRANCOISNicolas FRANCOIS
3,7771516
edited Jun 21 '18 at 5:58
Did
249k23227466
asked Jun 21 '18 at 2:16
Nicolas FRANCOISNicolas FRANCOIS
3,7771516
edited Jun 21 '18 at 5:58
Did
249k23227466
edited Jun 21 '18 at 5:58
Did
249k23227466
edited Jun 21 '18 at 5:58
Did
249k23227466
249k23227466
asked Jun 21 '18 at 2:16
Nicolas FRANCOISNicolas FRANCOIS
3,7771516
asked Jun 21 '18 at 2:16
Nicolas FRANCOISNicolas FRANCOIS
3,7771516
asked Jun 21 '18 at 2:16
Nicolas FRANCOISNicolas FRANCOIS
3,7771516
3,7771516
3
$begingroup$
In other words, is there a "natural" (whatever that means) explanation of the identity $$Eleft(frac1X+1right)=E(q^U)$$ for $X$ binomial $(n,p)$, $U$ uniform on $0,1,ldots,n$, and $q=1-p$? Interesting question...
$endgroup$
– Did
Jun 21 '18 at 5:55
$begingroup$
One can possibly justify the interchange of integral and summation to say that $Eleft(frac11+Xright)=Eleft(int_0^1 t^X,dtright)=int_0^1 E(t^X),dt=int_0^1 (1-p+pt)^n,dt$. But that's just another proof, not sure if this might be helpful in this regard.
$endgroup$
– StubbornAtom
Jun 21 '18 at 11:48
add a comment |
3
$begingroup$
In other words, is there a "natural" (whatever that means) explanation of the identity $$Eleft(frac1X+1right)=E(q^U)$$ for $X$ binomial $(n,p)$, $U$ uniform on $0,1,ldots,n$, and $q=1-p$? Interesting question...
$endgroup$
– Did
Jun 21 '18 at 5:55
$begingroup$
One can possibly justify the interchange of integral and summation to say that $Eleft(frac11+Xright)=Eleft(int_0^1 t^X,dtright)=int_0^1 E(t^X),dt=int_0^1 (1-p+pt)^n,dt$. But that's just another proof, not sure if this might be helpful in this regard.
$endgroup$
– StubbornAtom
Jun 21 '18 at 11:48
3
3
$begingroup$
In other words, is there a "natural" (whatever that means) explanation of the identity $$Eleft(frac1X+1right)=E(q^U)$$ for $X$ binomial $(n,p)$, $U$ uniform on $0,1,ldots,n$, and $q=1-p$? Interesting question...
$endgroup$
– Did
Jun 21 '18 at 5:55
$begingroup$
In other words, is there a "natural" (whatever that means) explanation of the identity $$Eleft(frac1X+1right)=E(q^U)$$ for $X$ binomial $(n,p)$, $U$ uniform on $0,1,ldots,n$, and $q=1-p$? Interesting question...
$endgroup$
– Did
Jun 21 '18 at 5:55
$begingroup$
One can possibly justify the interchange of integral and summation to say that $Eleft(frac11+Xright)=Eleft(int_0^1 t^X,dtright)=int_0^1 E(t^X),dt=int_0^1 (1-p+pt)^n,dt$. But that's just another proof, not sure if this might be helpful in this regard.
$endgroup$
– StubbornAtom
Jun 21 '18 at 11:48
$begingroup$
One can possibly justify the interchange of integral and summation to say that $Eleft(frac11+Xright)=Eleft(int_0^1 t^X,dtright)=int_0^1 E(t^X),dt=int_0^1 (1-p+pt)^n,dt$. But that's just another proof, not sure if this might be helpful in this regard.
$endgroup$
– StubbornAtom
Jun 21 '18 at 11:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here is an application of your question to a setting that highly interests
me. In queueing there is the notion of utilization which is the long-run
fraction of time a server is busy serving demand. Consider a Markovian service
setting where demand arrives according to Poisson with $lambda=1$
and there are $N+1$ servers with Exponential service time and mean rate $mu=1$, where $NsimtextBinleft(n,pright)$.
An application of this is the utilization of an Uber driver; the number
of Uber drivers on a given instance is uncertain as it cannot be mandated or enforced by
the firm. Given $k$ drivers choose to drive on the streets, their
utilization would be $fraclambdamucdot k=frac1k$. So,
in this setting if we assume the above model by the law of total probability the utilization of
an Uber driver would be $Eleft[frac11+Nright]$. Given this,
I would be interested to know of an interpretation of the RHS? Any
ideas?
answered Mar 29 at 15:23
Konstantinos I. StourasKonstantinos I. Stouras
436
$endgroup$
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is an application of your question to a setting that highly interests
me. In queueing there is the notion of utilization which is the long-run
fraction of time a server is busy serving demand. Consider a Markovian service
setting where demand arrives according to Poisson with $lambda=1$
and there are $N+1$ servers with Exponential service time and mean rate $mu=1$, where $NsimtextBinleft(n,pright)$.
An application of this is the utilization of an Uber driver; the number
of Uber drivers on a given instance is uncertain as it cannot be mandated or enforced by
the firm. Given $k$ drivers choose to drive on the streets, their
utilization would be $fraclambdamucdot k=frac1k$. So,
in this setting if we assume the above model by the law of total probability the utilization of
an Uber driver would be $Eleft[frac11+Nright]$. Given this,
I would be interested to know of an interpretation of the RHS? Any
ideas?
answered Mar 29 at 15:23
Konstantinos I. StourasKonstantinos I. Stouras
436
$endgroup$
add a comment |
$begingroup$
Here is an application of your question to a setting that highly interests
me. In queueing there is the notion of utilization which is the long-run
fraction of time a server is busy serving demand. Consider a Markovian service
setting where demand arrives according to Poisson with $lambda=1$
and there are $N+1$ servers with Exponential service time and mean rate $mu=1$, where $NsimtextBinleft(n,pright)$.
An application of this is the utilization of an Uber driver; the number
of Uber drivers on a given instance is uncertain as it cannot be mandated or enforced by
the firm. Given $k$ drivers choose to drive on the streets, their
utilization would be $fraclambdamucdot k=frac1k$. So,
in this setting if we assume the above model by the law of total probability the utilization of
an Uber driver would be $Eleft[frac11+Nright]$. Given this,
I would be interested to know of an interpretation of the RHS? Any
ideas?
answered Mar 29 at 15:23
Konstantinos I. StourasKonstantinos I. Stouras
436
$endgroup$
add a comment |
$begingroup$
Here is an application of your question to a setting that highly interests
me. In queueing there is the notion of utilization which is the long-run
fraction of time a server is busy serving demand. Consider a Markovian service
setting where demand arrives according to Poisson with $lambda=1$
and there are $N+1$ servers with Exponential service time and mean rate $mu=1$, where $NsimtextBinleft(n,pright)$.
An application of this is the utilization of an Uber driver; the number
of Uber drivers on a given instance is uncertain as it cannot be mandated or enforced by
the firm. Given $k$ drivers choose to drive on the streets, their
utilization would be $fraclambdamucdot k=frac1k$. So,
in this setting if we assume the above model by the law of total probability the utilization of
an Uber driver would be $Eleft[frac11+Nright]$. Given this,
I would be interested to know of an interpretation of the RHS? Any
ideas?
answered Mar 29 at 15:23
Konstantinos I. StourasKonstantinos I. Stouras
436
$endgroup$
Here is an application of your question to a setting that highly interests
me. In queueing there is the notion of utilization which is the long-run
fraction of time a server is busy serving demand. Consider a Markovian service
setting where demand arrives according to Poisson with $lambda=1$
and there are $N+1$ servers with Exponential service time and mean rate $mu=1$, where $NsimtextBinleft(n,pright)$.
An application of this is the utilization of an Uber driver; the number
of Uber drivers on a given instance is uncertain as it cannot be mandated or enforced by
the firm. Given $k$ drivers choose to drive on the streets, their
utilization would be $fraclambdamucdot k=frac1k$. So,
in this setting if we assume the above model by the law of total probability the utilization of
an Uber driver would be $Eleft[frac11+Nright]$. Given this,
I would be interested to know of an interpretation of the RHS? Any
ideas?
answered Mar 29 at 15:23
Konstantinos I. StourasKonstantinos I. Stouras
436
edited Mar 29 at 18:31
answered Mar 29 at 15:23
Konstantinos I. StourasKonstantinos I. Stouras
436
answered Mar 29 at 15:23
Konstantinos I. StourasKonstantinos I. Stouras
436
answered Mar 29 at 15:23
Konstantinos I. StourasKonstantinos I. Stouras
436
436
add a comment |
add a comment |
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3
$begingroup$
In other words, is there a "natural" (whatever that means) explanation of the identity $$Eleft(frac1X+1right)=E(q^U)$$ for $X$ binomial $(n,p)$, $U$ uniform on $0,1,ldots,n$, and $q=1-p$? Interesting question...
$endgroup$
– Did
Jun 21 '18 at 5:55
$begingroup$
One can possibly justify the interchange of integral and summation to say that $Eleft(frac11+Xright)=Eleft(int_0^1 t^X,dtright)=int_0^1 E(t^X),dt=int_0^1 (1-p+pt)^n,dt$. But that's just another proof, not sure if this might be helpful in this regard.
$endgroup$
– StubbornAtom
Jun 21 '18 at 11:48