Mixing
Is it possible to parameterize the model by $Psi = frac{1-theta}{theta}$ ? Prove your answer
$begingroup$
A random sample of $6$ observations $(X_1, X_2, cdots, X_6)$ is generated from a Geometric($theta$), where $theta in (0, 1)$ unknown, but only $T = sum_{i=1}^{6} X_i$ is observed by the statistician.
(a) Describe the statistical model for the observed data ($T$)
(b)-(i) Is it possible to parameterize the model by $Psi = frac{1-theta}{theta}$ ? Prove your answer
(b)-(ii) Is it possible to parameterize the model by $Psi = theta(1-theta)$ ? Prove your answer
My attempt:
(a) Since geometric is iid with Negative Binomial
Each $X_i$ ~ $mathrm{Geometric}(theta)$ therefore $T = sum_{i=1}^{6}$ ~ $mathrm{NegativeBinomial}(r, theta)$ where $theta in (0, 1)$ unknown.
The probability function for T is given by $$f_{theta}(t) = {t +r-1choose t}(1-theta)^t theta^r$$ for $t = 0,1, cdots, 6$
parameter is $theta$ and parameter space is $[0,1]$
(b)-(i) and (b)-(ii) I'm not sure how to do. Would I just show they are one to one by graphing each of them? Not sure
probability statistics
edited Jan 25 at 4:04
Lee David Chung Lin
4,39341242
asked Jan 25 at 3:32
Bas basBas bas
49012
$endgroup$
add a comment |
$begingroup$
A random sample of $6$ observations $(X_1, X_2, cdots, X_6)$ is generated from a Geometric($theta$), where $theta in (0, 1)$ unknown, but only $T = sum_{i=1}^{6} X_i$ is observed by the statistician.
(a) Describe the statistical model for the observed data ($T$)
(b)-(i) Is it possible to parameterize the model by $Psi = frac{1-theta}{theta}$ ? Prove your answer
(b)-(ii) Is it possible to parameterize the model by $Psi = theta(1-theta)$ ? Prove your answer
My attempt:
(a) Since geometric is iid with Negative Binomial
Each $X_i$ ~ $mathrm{Geometric}(theta)$ therefore $T = sum_{i=1}^{6}$ ~ $mathrm{NegativeBinomial}(r, theta)$ where $theta in (0, 1)$ unknown.
The probability function for T is given by $$f_{theta}(t) = {t +r-1choose t}(1-theta)^t theta^r$$ for $t = 0,1, cdots, 6$
parameter is $theta$ and parameter space is $[0,1]$
(b)-(i) and (b)-(ii) I'm not sure how to do. Would I just show they are one to one by graphing each of them? Not sure
probability statistics
edited Jan 25 at 4:04
Lee David Chung Lin
4,39341242
asked Jan 25 at 3:32
Bas basBas bas
49012
$endgroup$
add a comment |
$begingroup$
A random sample of $6$ observations $(X_1, X_2, cdots, X_6)$ is generated from a Geometric($theta$), where $theta in (0, 1)$ unknown, but only $T = sum_{i=1}^{6} X_i$ is observed by the statistician.
(a) Describe the statistical model for the observed data ($T$)
(b)-(i) Is it possible to parameterize the model by $Psi = frac{1-theta}{theta}$ ? Prove your answer
(b)-(ii) Is it possible to parameterize the model by $Psi = theta(1-theta)$ ? Prove your answer
My attempt:
(a) Since geometric is iid with Negative Binomial
Each $X_i$ ~ $mathrm{Geometric}(theta)$ therefore $T = sum_{i=1}^{6}$ ~ $mathrm{NegativeBinomial}(r, theta)$ where $theta in (0, 1)$ unknown.
The probability function for T is given by $$f_{theta}(t) = {t +r-1choose t}(1-theta)^t theta^r$$ for $t = 0,1, cdots, 6$
parameter is $theta$ and parameter space is $[0,1]$
(b)-(i) and (b)-(ii) I'm not sure how to do. Would I just show they are one to one by graphing each of them? Not sure
probability statistics
edited Jan 25 at 4:04
Lee David Chung Lin
4,39341242
asked Jan 25 at 3:32
Bas basBas bas
49012
$endgroup$
A random sample of $6$ observations $(X_1, X_2, cdots, X_6)$ is generated from a Geometric($theta$), where $theta in (0, 1)$ unknown, but only $T = sum_{i=1}^{6} X_i$ is observed by the statistician.
(a) Describe the statistical model for the observed data ($T$)
(b)-(i) Is it possible to parameterize the model by $Psi = frac{1-theta}{theta}$ ? Prove your answer
(b)-(ii) Is it possible to parameterize the model by $Psi = theta(1-theta)$ ? Prove your answer
My attempt:
(a) Since geometric is iid with Negative Binomial
Each $X_i$ ~ $mathrm{Geometric}(theta)$ therefore $T = sum_{i=1}^{6}$ ~ $mathrm{NegativeBinomial}(r, theta)$ where $theta in (0, 1)$ unknown.
The probability function for T is given by $$f_{theta}(t) = {t +r-1choose t}(1-theta)^t theta^r$$ for $t = 0,1, cdots, 6$
parameter is $theta$ and parameter space is $[0,1]$
(b)-(i) and (b)-(ii) I'm not sure how to do. Would I just show they are one to one by graphing each of them? Not sure
probability statistics
probability statistics
edited Jan 25 at 4:04
Lee David Chung Lin
4,39341242
asked Jan 25 at 3:32
Bas basBas bas
49012
edited Jan 25 at 4:04
Lee David Chung Lin
4,39341242
asked Jan 25 at 3:32
Bas basBas bas
49012
edited Jan 25 at 4:04
Lee David Chung Lin
4,39341242
edited Jan 25 at 4:04
Lee David Chung Lin
4,39341242
edited Jan 25 at 4:04
Lee David Chung Lin
4,39341242
4,39341242
asked Jan 25 at 3:32
Bas basBas bas
49012
asked Jan 25 at 3:32
Bas basBas bas
49012
asked Jan 25 at 3:32
Bas basBas bas
49012
49012
add a comment |
add a comment |
1 Answer
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$begingroup$
Your answer for (a) is a probability mass function so personally I would not use $f$ to describe it as that is often used for a density. I would write $6$ rather than $r$. And it depends on whether your negative binomial starts at $0$ or $6$, i.e. whether your geometric distributions start at $0$ or $1$
For (b)(i) you can say that $Psi = frac{1-theta}{theta} iff theta = frac{1}{Psi+1}$ at least if $theta not = 0$ which it has to be for the question to be sensible, so you can just use this as a substitution to parameterize the model by $Psi$
For (b)(ii) you can say that $Psi = theta({1-theta})$ is satisfied by $theta = frac{1 pm sqrt{1+4Psi}}{2}$, and the $pm$ is a warning sign; for example $Psi=0.09$ could be produced by $theta=0.1$ or $theta=0.9$ which would give you different probabilities, so so you cannot parameterize the model by this $Psi$
answered Jan 25 at 8:43
HenryHenry
101k482168
$endgroup$
$begingroup$
"And it depends on whether your negative binomial starts at 0 or 6, i.e. whether your geometric distributions start at 0 or 1". What do you mean? Confuse at this part get the rest though
$endgroup$
– Bas bas
Jan 25 at 13:37
$begingroup$
A negative binomial distribution may count the number of failures before the $n$th success (minimum possible value is $0$) or the number of attempts when then $n$th success occurs (minimum possible value is $n$). When $n=1$ you have a geometric distribution with a minimum possible value of $0$ or $1$ depending on which definition you are using; your question does not say.
$endgroup$
– Henry
Jan 25 at 14:30
add a comment |
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1 Answer
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$begingroup$
Your answer for (a) is a probability mass function so personally I would not use $f$ to describe it as that is often used for a density. I would write $6$ rather than $r$. And it depends on whether your negative binomial starts at $0$ or $6$, i.e. whether your geometric distributions start at $0$ or $1$
For (b)(i) you can say that $Psi = frac{1-theta}{theta} iff theta = frac{1}{Psi+1}$ at least if $theta not = 0$ which it has to be for the question to be sensible, so you can just use this as a substitution to parameterize the model by $Psi$
For (b)(ii) you can say that $Psi = theta({1-theta})$ is satisfied by $theta = frac{1 pm sqrt{1+4Psi}}{2}$, and the $pm$ is a warning sign; for example $Psi=0.09$ could be produced by $theta=0.1$ or $theta=0.9$ which would give you different probabilities, so so you cannot parameterize the model by this $Psi$
answered Jan 25 at 8:43
HenryHenry
101k482168
$endgroup$
$begingroup$
"And it depends on whether your negative binomial starts at 0 or 6, i.e. whether your geometric distributions start at 0 or 1". What do you mean? Confuse at this part get the rest though
$endgroup$
– Bas bas
Jan 25 at 13:37
$begingroup$
A negative binomial distribution may count the number of failures before the $n$th success (minimum possible value is $0$) or the number of attempts when then $n$th success occurs (minimum possible value is $n$). When $n=1$ you have a geometric distribution with a minimum possible value of $0$ or $1$ depending on which definition you are using; your question does not say.
$endgroup$
– Henry
Jan 25 at 14:30
add a comment |
$begingroup$
Your answer for (a) is a probability mass function so personally I would not use $f$ to describe it as that is often used for a density. I would write $6$ rather than $r$. And it depends on whether your negative binomial starts at $0$ or $6$, i.e. whether your geometric distributions start at $0$ or $1$
For (b)(i) you can say that $Psi = frac{1-theta}{theta} iff theta = frac{1}{Psi+1}$ at least if $theta not = 0$ which it has to be for the question to be sensible, so you can just use this as a substitution to parameterize the model by $Psi$
For (b)(ii) you can say that $Psi = theta({1-theta})$ is satisfied by $theta = frac{1 pm sqrt{1+4Psi}}{2}$, and the $pm$ is a warning sign; for example $Psi=0.09$ could be produced by $theta=0.1$ or $theta=0.9$ which would give you different probabilities, so so you cannot parameterize the model by this $Psi$
answered Jan 25 at 8:43
HenryHenry
101k482168
$endgroup$
$begingroup$
"And it depends on whether your negative binomial starts at 0 or 6, i.e. whether your geometric distributions start at 0 or 1". What do you mean? Confuse at this part get the rest though
$endgroup$
– Bas bas
Jan 25 at 13:37
$begingroup$
A negative binomial distribution may count the number of failures before the $n$th success (minimum possible value is $0$) or the number of attempts when then $n$th success occurs (minimum possible value is $n$). When $n=1$ you have a geometric distribution with a minimum possible value of $0$ or $1$ depending on which definition you are using; your question does not say.
$endgroup$
– Henry
Jan 25 at 14:30
add a comment |
$begingroup$
Your answer for (a) is a probability mass function so personally I would not use $f$ to describe it as that is often used for a density. I would write $6$ rather than $r$. And it depends on whether your negative binomial starts at $0$ or $6$, i.e. whether your geometric distributions start at $0$ or $1$
For (b)(i) you can say that $Psi = frac{1-theta}{theta} iff theta = frac{1}{Psi+1}$ at least if $theta not = 0$ which it has to be for the question to be sensible, so you can just use this as a substitution to parameterize the model by $Psi$
For (b)(ii) you can say that $Psi = theta({1-theta})$ is satisfied by $theta = frac{1 pm sqrt{1+4Psi}}{2}$, and the $pm$ is a warning sign; for example $Psi=0.09$ could be produced by $theta=0.1$ or $theta=0.9$ which would give you different probabilities, so so you cannot parameterize the model by this $Psi$
answered Jan 25 at 8:43
HenryHenry
101k482168
$endgroup$
Your answer for (a) is a probability mass function so personally I would not use $f$ to describe it as that is often used for a density. I would write $6$ rather than $r$. And it depends on whether your negative binomial starts at $0$ or $6$, i.e. whether your geometric distributions start at $0$ or $1$
For (b)(i) you can say that $Psi = frac{1-theta}{theta} iff theta = frac{1}{Psi+1}$ at least if $theta not = 0$ which it has to be for the question to be sensible, so you can just use this as a substitution to parameterize the model by $Psi$
For (b)(ii) you can say that $Psi = theta({1-theta})$ is satisfied by $theta = frac{1 pm sqrt{1+4Psi}}{2}$, and the $pm$ is a warning sign; for example $Psi=0.09$ could be produced by $theta=0.1$ or $theta=0.9$ which would give you different probabilities, so so you cannot parameterize the model by this $Psi$
answered Jan 25 at 8:43
HenryHenry
101k482168
answered Jan 25 at 8:43
HenryHenry
101k482168
answered Jan 25 at 8:43
HenryHenry
101k482168
answered Jan 25 at 8:43
HenryHenry
101k482168
101k482168
$begingroup$
"And it depends on whether your negative binomial starts at 0 or 6, i.e. whether your geometric distributions start at 0 or 1". What do you mean? Confuse at this part get the rest though
$endgroup$
– Bas bas
Jan 25 at 13:37
$begingroup$
A negative binomial distribution may count the number of failures before the $n$th success (minimum possible value is $0$) or the number of attempts when then $n$th success occurs (minimum possible value is $n$). When $n=1$ you have a geometric distribution with a minimum possible value of $0$ or $1$ depending on which definition you are using; your question does not say.
$endgroup$
– Henry
Jan 25 at 14:30
add a comment |
$begingroup$
"And it depends on whether your negative binomial starts at 0 or 6, i.e. whether your geometric distributions start at 0 or 1". What do you mean? Confuse at this part get the rest though
$endgroup$
– Bas bas
Jan 25 at 13:37
$begingroup$
A negative binomial distribution may count the number of failures before the $n$th success (minimum possible value is $0$) or the number of attempts when then $n$th success occurs (minimum possible value is $n$). When $n=1$ you have a geometric distribution with a minimum possible value of $0$ or $1$ depending on which definition you are using; your question does not say.
$endgroup$
– Henry
Jan 25 at 14:30
$begingroup$
"And it depends on whether your negative binomial starts at 0 or 6, i.e. whether your geometric distributions start at 0 or 1". What do you mean? Confuse at this part get the rest though
$endgroup$
– Bas bas
Jan 25 at 13:37
$begingroup$
"And it depends on whether your negative binomial starts at 0 or 6, i.e. whether your geometric distributions start at 0 or 1". What do you mean? Confuse at this part get the rest though
$endgroup$
– Bas bas
Jan 25 at 13:37
$begingroup$
A negative binomial distribution may count the number of failures before the $n$th success (minimum possible value is $0$) or the number of attempts when then $n$th success occurs (minimum possible value is $n$). When $n=1$ you have a geometric distribution with a minimum possible value of $0$ or $1$ depending on which definition you are using; your question does not say.
$endgroup$
– Henry
Jan 25 at 14:30
$begingroup$
A negative binomial distribution may count the number of failures before the $n$th success (minimum possible value is $0$) or the number of attempts when then $n$th success occurs (minimum possible value is $n$). When $n=1$ you have a geometric distribution with a minimum possible value of $0$ or $1$ depending on which definition you are using; your question does not say.
$endgroup$
– Henry
Jan 25 at 14:30
add a comment |
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