Log-likelihood of zero-truncated Poisson
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Question:
Edwards and Eberhardt (1967) conducted a live-trapping study on a confined population of known size. In their study, wild cottontail
rabbits were penned in a 4-acre rabbit-proof enclosure. Live trapping was conducted for 18 consecutive
nights. Recorded capture frequencies were as follows: 43 rabbits were caught at exactly 1 night, 16
rabbits at 2 nights, 8 at 3 nights, 6 at four nights, 2 at 6 nights and 1 rabbit was caught at 7 nights. In
other words, we have observations from $X_1, · · · X_m$ count variables of m = 76 different, caught rabbits.
So, in this sample, 43 of the $x_1, · · · , x_m$ were ones, 16 twos, etc. Every count can take a maximum of
T = 18 captures. The number of of zero counts, i.e. the number of rabbits that were not caught, is
unobserved. We denote this additional sample as $X_{m+1}, · · · , X_n$. Assuming independence of trapping
occasions and homogeneity of capture probability θ at each occasion each $X_i$ would follow a binomial distribution. We assume here that a Poisson approximation is justified. That is:
$P(X_i = x) = frac{exp(−θT)(θt)^x}{x!}$ where i = 1, · · · , n.
Note that the observed sample is zero-truncated so that the observed sample can be considered as arising from the zero-truncated Poisson density:
$$exp(−θT)(θT)^x over (1 − exp(−θT))x!$$ for $i = 1, · · · , m$, $;θ ∈ (0, 1)$ and $x ∈ {0, 1, · · · , T}$.
Determine the log-likelihood function for the observed sample.
My solution:
I have already attempted this and got:
$sum_{i=1}^m n_ilogfrac{exp(−θT)(θT)^x}{x!}frac{1}{1 − exp(−θT)}$
Is this correct? If not, what should I have got? What have I done wrong?
poisson-distribution log-likelihood
asked Nov 21 at 22:34
Vicky
597
add a comment |
up vote
1
down vote
favorite
Question:
Edwards and Eberhardt (1967) conducted a live-trapping study on a confined population of known size. In their study, wild cottontail
rabbits were penned in a 4-acre rabbit-proof enclosure. Live trapping was conducted for 18 consecutive
nights. Recorded capture frequencies were as follows: 43 rabbits were caught at exactly 1 night, 16
rabbits at 2 nights, 8 at 3 nights, 6 at four nights, 2 at 6 nights and 1 rabbit was caught at 7 nights. In
other words, we have observations from $X_1, · · · X_m$ count variables of m = 76 different, caught rabbits.
So, in this sample, 43 of the $x_1, · · · , x_m$ were ones, 16 twos, etc. Every count can take a maximum of
T = 18 captures. The number of of zero counts, i.e. the number of rabbits that were not caught, is
unobserved. We denote this additional sample as $X_{m+1}, · · · , X_n$. Assuming independence of trapping
occasions and homogeneity of capture probability θ at each occasion each $X_i$ would follow a binomial distribution. We assume here that a Poisson approximation is justified. That is:
$P(X_i = x) = frac{exp(−θT)(θt)^x}{x!}$ where i = 1, · · · , n.
Note that the observed sample is zero-truncated so that the observed sample can be considered as arising from the zero-truncated Poisson density:
$$exp(−θT)(θT)^x over (1 − exp(−θT))x!$$ for $i = 1, · · · , m$, $;θ ∈ (0, 1)$ and $x ∈ {0, 1, · · · , T}$.
Determine the log-likelihood function for the observed sample.
My solution:
I have already attempted this and got:
$sum_{i=1}^m n_ilogfrac{exp(−θT)(θT)^x}{x!}frac{1}{1 − exp(−θT)}$
Is this correct? If not, what should I have got? What have I done wrong?
poisson-distribution log-likelihood
asked Nov 21 at 22:34
Vicky
597
Someone's been doing the Dankmar's coursework here :P I think the context is not relevant anyway, you can just ask for the log-likelihood function of a zero-truncated Poisson distribution and its first and second order derivative
– Euler_Salter
Nov 21 at 22:42
@Euler_Salter haha yes! I thought context was good. Does it look right to you?
– Vicky
Nov 21 at 22:49
Yes but it could scare off lazy fellow stackexchangers, so it's nice to keep it short! Not sure now, I'm a bit tired, but I just found this: math.stackexchange.com/a/1857529/348937
– Euler_Salter
Nov 21 at 22:51
@Euler_Salter thanks! Will take a look.
– Vicky
Nov 21 at 22:53
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Question:
Edwards and Eberhardt (1967) conducted a live-trapping study on a confined population of known size. In their study, wild cottontail
rabbits were penned in a 4-acre rabbit-proof enclosure. Live trapping was conducted for 18 consecutive
nights. Recorded capture frequencies were as follows: 43 rabbits were caught at exactly 1 night, 16
rabbits at 2 nights, 8 at 3 nights, 6 at four nights, 2 at 6 nights and 1 rabbit was caught at 7 nights. In
other words, we have observations from $X_1, · · · X_m$ count variables of m = 76 different, caught rabbits.
So, in this sample, 43 of the $x_1, · · · , x_m$ were ones, 16 twos, etc. Every count can take a maximum of
T = 18 captures. The number of of zero counts, i.e. the number of rabbits that were not caught, is
unobserved. We denote this additional sample as $X_{m+1}, · · · , X_n$. Assuming independence of trapping
occasions and homogeneity of capture probability θ at each occasion each $X_i$ would follow a binomial distribution. We assume here that a Poisson approximation is justified. That is:
$P(X_i = x) = frac{exp(−θT)(θt)^x}{x!}$ where i = 1, · · · , n.
Note that the observed sample is zero-truncated so that the observed sample can be considered as arising from the zero-truncated Poisson density:
$$exp(−θT)(θT)^x over (1 − exp(−θT))x!$$ for $i = 1, · · · , m$, $;θ ∈ (0, 1)$ and $x ∈ {0, 1, · · · , T}$.
Determine the log-likelihood function for the observed sample.
My solution:
I have already attempted this and got:
$sum_{i=1}^m n_ilogfrac{exp(−θT)(θT)^x}{x!}frac{1}{1 − exp(−θT)}$
Is this correct? If not, what should I have got? What have I done wrong?
poisson-distribution log-likelihood
asked Nov 21 at 22:34
Vicky
597
Question:
Edwards and Eberhardt (1967) conducted a live-trapping study on a confined population of known size. In their study, wild cottontail
rabbits were penned in a 4-acre rabbit-proof enclosure. Live trapping was conducted for 18 consecutive
nights. Recorded capture frequencies were as follows: 43 rabbits were caught at exactly 1 night, 16
rabbits at 2 nights, 8 at 3 nights, 6 at four nights, 2 at 6 nights and 1 rabbit was caught at 7 nights. In
other words, we have observations from $X_1, · · · X_m$ count variables of m = 76 different, caught rabbits.
So, in this sample, 43 of the $x_1, · · · , x_m$ were ones, 16 twos, etc. Every count can take a maximum of
T = 18 captures. The number of of zero counts, i.e. the number of rabbits that were not caught, is
unobserved. We denote this additional sample as $X_{m+1}, · · · , X_n$. Assuming independence of trapping
occasions and homogeneity of capture probability θ at each occasion each $X_i$ would follow a binomial distribution. We assume here that a Poisson approximation is justified. That is:
$P(X_i = x) = frac{exp(−θT)(θt)^x}{x!}$ where i = 1, · · · , n.
Note that the observed sample is zero-truncated so that the observed sample can be considered as arising from the zero-truncated Poisson density:
$$exp(−θT)(θT)^x over (1 − exp(−θT))x!$$ for $i = 1, · · · , m$, $;θ ∈ (0, 1)$ and $x ∈ {0, 1, · · · , T}$.
Determine the log-likelihood function for the observed sample.
My solution:
I have already attempted this and got:
$sum_{i=1}^m n_ilogfrac{exp(−θT)(θT)^x}{x!}frac{1}{1 − exp(−θT)}$
Is this correct? If not, what should I have got? What have I done wrong?
poisson-distribution log-likelihood
poisson-distribution log-likelihood
asked Nov 21 at 22:34
Vicky
597
asked Nov 21 at 22:34
Vicky
597
asked Nov 21 at 22:34
Vicky
597
asked Nov 21 at 22:34
Vicky
597
asked Nov 21 at 22:34
Vicky
597
597
Someone's been doing the Dankmar's coursework here :P I think the context is not relevant anyway, you can just ask for the log-likelihood function of a zero-truncated Poisson distribution and its first and second order derivative
– Euler_Salter
Nov 21 at 22:42
@Euler_Salter haha yes! I thought context was good. Does it look right to you?
– Vicky
Nov 21 at 22:49
Yes but it could scare off lazy fellow stackexchangers, so it's nice to keep it short! Not sure now, I'm a bit tired, but I just found this: math.stackexchange.com/a/1857529/348937
– Euler_Salter
Nov 21 at 22:51
@Euler_Salter thanks! Will take a look.
– Vicky
Nov 21 at 22:53
add a comment |
Someone's been doing the Dankmar's coursework here :P I think the context is not relevant anyway, you can just ask for the log-likelihood function of a zero-truncated Poisson distribution and its first and second order derivative
– Euler_Salter
Nov 21 at 22:42
@Euler_Salter haha yes! I thought context was good. Does it look right to you?
– Vicky
Nov 21 at 22:49
Yes but it could scare off lazy fellow stackexchangers, so it's nice to keep it short! Not sure now, I'm a bit tired, but I just found this: math.stackexchange.com/a/1857529/348937
– Euler_Salter
Nov 21 at 22:51
@Euler_Salter thanks! Will take a look.
– Vicky
Nov 21 at 22:53
Someone's been doing the Dankmar's coursework here :P I think the context is not relevant anyway, you can just ask for the log-likelihood function of a zero-truncated Poisson distribution and its first and second order derivative
– Euler_Salter
Nov 21 at 22:42
Someone's been doing the Dankmar's coursework here :P I think the context is not relevant anyway, you can just ask for the log-likelihood function of a zero-truncated Poisson distribution and its first and second order derivative
– Euler_Salter
Nov 21 at 22:42
@Euler_Salter haha yes! I thought context was good. Does it look right to you?
– Vicky
Nov 21 at 22:49
@Euler_Salter haha yes! I thought context was good. Does it look right to you?
– Vicky
Nov 21 at 22:49
Yes but it could scare off lazy fellow stackexchangers, so it's nice to keep it short! Not sure now, I'm a bit tired, but I just found this: math.stackexchange.com/a/1857529/348937
– Euler_Salter
Nov 21 at 22:51
Yes but it could scare off lazy fellow stackexchangers, so it's nice to keep it short! Not sure now, I'm a bit tired, but I just found this: math.stackexchange.com/a/1857529/348937
– Euler_Salter
Nov 21 at 22:51
@Euler_Salter thanks! Will take a look.
– Vicky
Nov 21 at 22:53
@Euler_Salter thanks! Will take a look.
– Vicky
Nov 21 at 22:53
add a comment |
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Someone's been doing the Dankmar's coursework here :P I think the context is not relevant anyway, you can just ask for the log-likelihood function of a zero-truncated Poisson distribution and its first and second order derivative
– Euler_Salter
Nov 21 at 22:42
@Euler_Salter haha yes! I thought context was good. Does it look right to you?
– Vicky
Nov 21 at 22:49
Yes but it could scare off lazy fellow stackexchangers, so it's nice to keep it short! Not sure now, I'm a bit tired, but I just found this: math.stackexchange.com/a/1857529/348937
– Euler_Salter
Nov 21 at 22:51
@Euler_Salter thanks! Will take a look.
– Vicky
Nov 21 at 22:53