1
$\begingroup$

This is a model that is used to model soccer scores, so $i$ and $j$ are, respectively, home and away teams. Random variables $(x,y)$ are the goals scored by the home and away teams, respectively. Parameter $\lambda$ is a known mean goals scored by the home team and $\mu$ is the mean goals scored by the away team. I have managed to fix all the other parameters except for $\rho$, which I have to estimate via MLE.

$$Pr(X_{i,j}=x, Y_{i,j}=y)=\tau_{\lambda, \mu}(x,y)\frac{\lambda^x \text{exp}(-\lambda)}{x!}\frac{\mu^y\text{exp}(-\mu)}{y!}$$ where $$\lambda=\alpha_{i}\beta_{j}\gamma$$ $$\mu=\alpha_{j}\beta_{i}$$ and $$\tau_{\lambda,\mu}(x,y)=\left\{\begin{array}{cc} 1-\lambda\mu\rho &\text{if $x=y=0$,} \\ 1+\lambda\rho &\text{if $x=0,y=1$,}\\ 1+\mu\rho &\text{if $x=1,y=0$,}\\ 1-\rho &\text{if $x=y=1$,}\\ 1 &\text{otherwise}\end{array} \right.$$

Based on the above equations, all the parameters $(\lambda, \mu, \alpha, \beta, \gamma)$ are known constants.

So, now, the problem that I am having is that I have no clue on how to estimate $\rho$ using the maximum likelihood function since a piece-wise equation is involved.

Also, it will be great if anyone can do this using R.

$\endgroup$
12
  • $\begingroup$ it is actually a Poisson distribution attached at the back, maybe that will clarify the problem. Sorry. $\endgroup$
    – SooBin
    Oct 31, 2013 at 16:07
  • $\begingroup$ This is a model that i am using to model soccer scores, so i and j are home and away team that are playing respectively. x and y are the goals scored by the home and away team respectively. I have managed to fix all the other parameters except for p which i have to estimate it via MLE. $\endgroup$
    – SooBin
    Oct 31, 2013 at 16:12
  • $\begingroup$ yeah, because like what you said, the probability distribution will be strange if i dont add that in. I was thinking that it was not necessary since it will be ignored when differentiating as it is independent of p. Sorry Zen. $\endgroup$
    – SooBin
    Oct 31, 2013 at 16:16
  • 1
    $\begingroup$ I've extensively edited your question for syntax and to add LATEX rendering for your mathematics. Please check that I have not accidentally introduced any errors. $\endgroup$
    – Sycorax
    Nov 1, 2013 at 2:45
  • 1
    $\begingroup$ I don't see anything "piecewise" about the $\tau$ at all: $\tau_{\lambda, \mu}(x,y)=\rho (-\lambda \mu (1-x) (1-y)+\lambda (1-x) y+\mu x (1-y))+1.$ Besides, $x$ and $y$ are data, not parameters, so all that matters is how $\tau$ depends on the parameters $\rho, \mu, \lambda$, and it's a simple (multilinear) polynomial in them (and linear in $\rho$). $\endgroup$
    – whuber
    Nov 1, 2013 at 20:03

1 Answer 1

1
$\begingroup$

If I understand it, your data is $\{(i_m,j_m,x_m,y_m) : m=1,\dots,n\}$, in which, for the $m$-th match, $i_m$ is the index of the home team, $j_m$ is the index of the away team ($i_m\neq j_m$), $x_m$ is the number of goals scored by the home team, and $y_m$ is the number of goals scored by the away team.

If we have $t$ teams disputing the $n$ matches, the likelihood $L(\rho,\gamma,\alpha_i,\beta_i, i=1,\dots,t)$ is proportional to $$ \prod_{m=1}^n \tau_{\lambda_m,\mu_m}(x_m,y_m)\,\lambda_m^{x_m} \,\mu_m^{y_m}\,e^{-(\lambda_m+\mu_m)} \, , $$ in which $\lambda_m = \alpha_{i_m} \beta_{j_m}\gamma$ and $\mu_m = \alpha_{i_m} \beta_{j_m}$. This likelihood is a function of $2(t+1)$ parameters. You will probably need to implement some conjugate gradient method to solve this constrained optimization problem. I wonder what happens if we put some priors on the parameters and MCMC the posterior with a random walk Metropolis. You said that you know the values of the parameters, except for $\rho$ (and I wonder how...). Hence, your first step is to write an R function that computes the product above for each value of $\rho$. Then, plot it and use the optimizer to see what happens.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.