We are given the additional information that conditional on the parameter of interest \lambda, our observations X_1, X_2, \ldots, X_ n are independently and identically distributed according to the probability distribution \textsf{Poiss}(\lambda ). From here, we compute our likelihood function. Our approach would be to compute a single function of \lambda L(X_ i | \lambda ), then plug in our data X_1, X_2, \ldots, X_ n to compute the overall likelihood L_ n(X_1, X_2, \ldots ,X_ n | \lambda ) = L(X_1 | \lambda )L(X_2|\lambda )\ldots L(X_ n|\lambda ).

Question

We are given the additional information that conditional on the parameter of interest \lambda, our observations X_1, X_2, \ldots, X_ n are independently and identically distributed according to the probability distribution \textsf{Poiss}(\lambda ). From here, we compute our likelihood function. Our approach would be to compute a single function of \lambda L(X_ i | \lambda ), then plug in our data X_1, X_2, \ldots, X_ n to compute the overall likelihood L_ n(X_1, X_2, \ldots ,X_ n | \lambda ) = L(X_1 | \lambda )L(X_2|\lambda )\ldots L(X_ n|\lambda ).

In our framework so far, we treat the observations X_1, \ldots , X_ n as fixed values, by which we perform Bayesian inference. Hence, L(X_ i | \lambda ) is to be viewed as a function of \lambda with X_ i as a parameter. Thus when using proportionality notation, we only need to consider variations concerning our parameter of interest \lambda. Compute the likelihood function L(X_1|\lambda ), using proportionality notation to simplify it such that in your expression, L(X_1|\lambda =1)=e^{-1} regardless of the value of X_1. (Note that this is not necessarily the actual likelihood L.)

Use X1 for X_1.

L(X_1|\lambda ) \propto

unanswered

Multiply the expressions L(X_1|\lambda ), L(X_2|\lambda ), \ldots, L(X_ n|\lambda ) based on the simplified expression for L(X_1|\lambda ) to get the desired likelihood function L_ n(X_1, X_2, \ldots ,X_ n | \lambda ).

Use SumXi for \sum _{i=1}^ n X_ i.

L_ n(X_1, X_2, \ldots ,X_ n | \lambda ) \propto

Answer 1

L_ n(X_1, X_2, \ldots ,X_ n | \lambda ) \propto e^{-X_1} e^{-X_2} \ldots e^{-X_ n} = e^{-(X_1 + X_2 + \ldots + X_ n)} = e^{-\text{SumXi}}

Answer 2

To compute the likelihood function for a Poisson distribution, we start by expressing L(X_1 | λ) using proportionality notation.

The probability mass function of the Poisson distribution is given by:

P(X_1 = x | λ) = (e^(-λ) * λ^x) / x!

Plugging in λ = 1, we have:

P(X_1 = x | λ = 1) = (e^(-1) * 1^x) / x!

Since we want L(X_1 | λ = 1) = e^(-1) regardless of the value of X_1, we set x = 1:

P(X_1 = 1 | λ = 1) = (e^(-1) * 1) / 1! = e^(-1)

So, the expression for L(X_1 | λ) in proportionality notation is:

L(X_1 | λ) ∝ e^(-1)

Now, to obtain the likelihood function for the entire dataset, we multiply the likelihoods for each observation:

L_n(X_1, X_2, ..., X_n | λ) ∝ L(X_1 | λ) * L(X_2 | λ) * ... * L(X_n | λ)

Using the simplified expression for L(X_1 | λ), we have:

L_n(X_1, X_2, ..., X_n | λ) ∝ (e^(-1))^n = e^(-n)

Therefore, the desired likelihood function in proportionality notation is:

L_n(X_1, X_2, ..., X_n | λ) ∝ e^(-n)

Note that this is a simplified expression and not necessarily the actual likelihood L.