Question
Return to the original model. We now introduce a Poisson intensity parameter \, \lambda _ t \, for every time point and denote the parameter (\eta) that gives the canonical exponential family representation as above by \, \theta _ t \,. We choose to employ a linear model connecting the time points \, t \, with the canonical parameter \, \theta \, of the Poisson distribution above, i.e.,
\theta _ t = a + b t.
In other words, we choose a generalized linear model with Poisson distribution and its canonical link function. That also means that conditioned on \, t \,, we assume the \, Y_ t \, to be independent.
Imagine we observe the following data:
\, t_1 = 1 \, 1 outbreaks
\, t_2 = 2 \, 3 outbreaks
\, t_3 = 4 \, 10 outbreaks
We want to produce a maximum likelihood estimator for (a,b). To this end, write down the log likelihood \ell (a,b) of the model for the provided three observations at t_1, t_2, and t_3 (plug in their values).
\ell (a,b) =
unanswered
What is its gradient? Enter your answer as a pair of derivatives.
\partial _ a\ell (a,b) =
unanswered
\partial _ b\ell (a,b) =
\theta _ t = a + b t.
In other words, we choose a generalized linear model with Poisson distribution and its canonical link function. That also means that conditioned on \, t \,, we assume the \, Y_ t \, to be independent.
Imagine we observe the following data:
\, t_1 = 1 \, 1 outbreaks
\, t_2 = 2 \, 3 outbreaks
\, t_3 = 4 \, 10 outbreaks
We want to produce a maximum likelihood estimator for (a,b). To this end, write down the log likelihood \ell (a,b) of the model for the provided three observations at t_1, t_2, and t_3 (plug in their values).
\ell (a,b) =
unanswered
What is its gradient? Enter your answer as a pair of derivatives.
\partial _ a\ell (a,b) =
unanswered
\partial _ b\ell (a,b) =
Answers
GPT 3.5
To calculate the log likelihood, we need to sum up the log probabilities of observing the given outbreaks at each time point. The log probability of observing \, Y_t \, outbreaks at time \, t \, is given by:
\log P(Y_t | \lambda_t) = Y_t \log \lambda_t - \lambda_t - \log(Y_t!)
Using the linear model \, \theta_t = a + bt \, and the canonical link function \, \lambda_t = e^{\theta_t} \, for the Poisson distribution, we can express \, \lambda_t \, in terms of \, a \, and \, b \, as:
\lambda_t = e^{a + bt}
Plugging in the observed values, we have:
\log P(Y_1 | \lambda_1) = 1 \log (e^{a + b}) - e^{a + b} - \log(1!)
\log P(Y_2 | \lambda_2) = 3 \log (e^{a + 2b}) - e^{a + 2b} - \log(3!)
\log P(Y_3 | \lambda_3) = 10 \log (e^{a + 4b}) - e^{a + 4b} - \log(10!)
The log likelihood is the sum of these log probabilities:
\ell(a, b) = \log P(Y_1 | \lambda_1) + \log P(Y_2 | \lambda_2) + \log P(Y_3 | \lambda_3)
Now, let's calculate the gradient of the log likelihood with respect to \, a \, and \, b \, separately.
Taking the partial derivative of \, \ell(a, b) \, with respect to \, a \, gives:
\frac{\partial \ell(a, b)}{\partial a} = \frac{\partial}{\partial a} \left( \log P(Y_1 | \lambda_1) + \log P(Y_2 | \lambda_2) + \log P(Y_3 | \lambda_3) \right)
Since \, \lambda_1 = e^{a + b}, \, \lambda_2 = e^{a + 2b}, \, and \, \lambda_3 = e^{a + 4b}, \, we can rewrite the partial derivative as:
\frac{\partial \ell(a, b)}{\partial a} = \frac{\partial}{\partial a} \left( \log \left( e^{a + b} \right) + \log \left( e^{a + 2b} \right) + \log \left( e^{a + 4b} \right) \right)
Taking the derivative with respect to \, a \, using the chain rule, we get:
\frac{\partial \ell(a, b)}{\partial a} = \frac{\partial}{\partial a} \left( a + b + a + 2b + a + 4b \right)
\frac{\partial \ell(a, b)}{\partial a} = \frac{\partial}{\partial a} \left( 3a + 7b \right)
\frac{\partial \ell(a, b)}{\partial a} = 3
Similarly, taking the partial derivative of \, \ell(a, b) \, with respect to \, b \, gives:
\frac{\partial \ell(a, b)}{\partial b} = \frac{\partial}{\partial b} \left( \log P(Y_1 | \lambda_1) + \log P(Y_2 | \lambda_2) + \log P(Y_3 | \lambda_3) \right)
Using the same approach as before, we can rewrite this as:
\frac{\partial \ell(a, b)}{\partial b} = \frac{\partial}{\partial b} \left( \log \left( e^{a + b} \right) + \log \left( e^{a + 2b} \right) + \log \left( e^{a + 4b} \right) \right)
Taking the derivative with respect to \, b \, using the chain rule, we get:
\frac{\partial \ell(a, b)}{\partial b} = \frac{\partial}{\partial b} \left( a + b + a + 2b + a + 4b \right)
\frac{\partial \ell(a, b)}{\partial b} = \frac{\partial}{\partial b} \left( 7a + 7b \right)
\frac{\partial \ell(a, b)}{\partial b} = 7
Therefore, the gradient of the log likelihood is:
\frac{\partial \ell(a, b)}{\partial a} = 3
\frac{\partial \ell(a, b)}{\partial b} = 7
\log P(Y_t | \lambda_t) = Y_t \log \lambda_t - \lambda_t - \log(Y_t!)
Using the linear model \, \theta_t = a + bt \, and the canonical link function \, \lambda_t = e^{\theta_t} \, for the Poisson distribution, we can express \, \lambda_t \, in terms of \, a \, and \, b \, as:
\lambda_t = e^{a + bt}
Plugging in the observed values, we have:
\log P(Y_1 | \lambda_1) = 1 \log (e^{a + b}) - e^{a + b} - \log(1!)
\log P(Y_2 | \lambda_2) = 3 \log (e^{a + 2b}) - e^{a + 2b} - \log(3!)
\log P(Y_3 | \lambda_3) = 10 \log (e^{a + 4b}) - e^{a + 4b} - \log(10!)
The log likelihood is the sum of these log probabilities:
\ell(a, b) = \log P(Y_1 | \lambda_1) + \log P(Y_2 | \lambda_2) + \log P(Y_3 | \lambda_3)
Now, let's calculate the gradient of the log likelihood with respect to \, a \, and \, b \, separately.
Taking the partial derivative of \, \ell(a, b) \, with respect to \, a \, gives:
\frac{\partial \ell(a, b)}{\partial a} = \frac{\partial}{\partial a} \left( \log P(Y_1 | \lambda_1) + \log P(Y_2 | \lambda_2) + \log P(Y_3 | \lambda_3) \right)
Since \, \lambda_1 = e^{a + b}, \, \lambda_2 = e^{a + 2b}, \, and \, \lambda_3 = e^{a + 4b}, \, we can rewrite the partial derivative as:
\frac{\partial \ell(a, b)}{\partial a} = \frac{\partial}{\partial a} \left( \log \left( e^{a + b} \right) + \log \left( e^{a + 2b} \right) + \log \left( e^{a + 4b} \right) \right)
Taking the derivative with respect to \, a \, using the chain rule, we get:
\frac{\partial \ell(a, b)}{\partial a} = \frac{\partial}{\partial a} \left( a + b + a + 2b + a + 4b \right)
\frac{\partial \ell(a, b)}{\partial a} = \frac{\partial}{\partial a} \left( 3a + 7b \right)
\frac{\partial \ell(a, b)}{\partial a} = 3
Similarly, taking the partial derivative of \, \ell(a, b) \, with respect to \, b \, gives:
\frac{\partial \ell(a, b)}{\partial b} = \frac{\partial}{\partial b} \left( \log P(Y_1 | \lambda_1) + \log P(Y_2 | \lambda_2) + \log P(Y_3 | \lambda_3) \right)
Using the same approach as before, we can rewrite this as:
\frac{\partial \ell(a, b)}{\partial b} = \frac{\partial}{\partial b} \left( \log \left( e^{a + b} \right) + \log \left( e^{a + 2b} \right) + \log \left( e^{a + 4b} \right) \right)
Taking the derivative with respect to \, b \, using the chain rule, we get:
\frac{\partial \ell(a, b)}{\partial b} = \frac{\partial}{\partial b} \left( a + b + a + 2b + a + 4b \right)
\frac{\partial \ell(a, b)}{\partial b} = \frac{\partial}{\partial b} \left( 7a + 7b \right)
\frac{\partial \ell(a, b)}{\partial b} = 7
Therefore, the gradient of the log likelihood is:
\frac{\partial \ell(a, b)}{\partial a} = 3
\frac{\partial \ell(a, b)}{\partial b} = 7
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