# Estimating latent Gaussian variable with maximum likelihood of Gaussian-product integral

## 2018/12/22

Suppose we have a population with a Normal distribution of some property, $t$, i.e. $t_i \sim N(\mu, \sigma_t$. For each member of the population we make a noisy measurement of the value, where the error distribution is Normal, $\epsilon_i \sim N(0, \sigma_i)$. The probability to make an observation $x_i$ is then,

$$P(x_i | t) ~ \sim \frac{1}{2 \pi \sigma_i \sigma_t} \int d\mu \exp{(-\frac{1}{2} \frac{(x_i - \mu)^2}{\sigma_i^2})} \exp{(-\frac{1}{2} \frac{(\mu - t)^2}{\sigma_t^2})}$$

In the last post I verified that the value of the integral is a Normal distribution,

$$P(x_i|t) \sim N(t, \sqrt{\sigma_i^2 + \sigma_t^2})$$

$$P(x_i|t) = \frac{1}{\sqrt{2 \pi (\sigma_i^2 + \sigma_t^2)}} \exp{(-\frac{1}{2} \frac{(x_i - t)^2}{\sigma_i^2 +\sigma_t^2})}$$

## Estimate of population parameters with maximum likelihood

Given a set of observations of $X$, we can use the above expression to estimate the population parameters, $t$ and $\sigma_t$. In this case the overall likelihood is the product of the individual likelihoods,

$$L = \Pi_i P(x_i | t)$$ Then the log-likelihood (actually -2 times the log-likelihood) is,

$$l = - 2 \ln{L} = \sum_i \frac{(x_i - t)^2}{\sigma_i^2 +\sigma_t^2} + \sum_i \ln{(\sigma_i^2 +\sigma_t^2)}$$

The maximum likelihood values are determined by setting the first derivatives to 0 and solving for $t$ and $\sigma_t$,

## value of t

$$\frac{\partial l}{\partial t} = 2 \sum_i \frac{(t - x_i)}{\sigma_i^2 +\sigma_t^2}$$

$$t = \frac{ \sum_i {x_i / (\sigma_i^2 +\sigma_t^2}) } {\sum_i {1 / (\sigma_i^2 +\sigma_t^2)} }$$

## value of $\sigma_t$

$$\frac{\partial l}{\partial \sigma_t} = 2 \sigma_t ( - \sum_i \frac{(t - x_i)^2}{(\sigma_i^2 +\sigma_t^2)^2} + \sum_i {\frac{1}{\sigma_i^2 +\sigma_t^2}})$$

## solving the equations

These equations are coupled and non-linear so there’s no closed form solution. However, we can investigate the limiting behavior in two interesting scenarios.

### All $\sigma_i$ equal

If $\sigma_i = \sigma_x$ for all $i$, then these simplify to

$$t = \sum_i x_i$$

$$\frac{1}{(\sigma_x^2 +\sigma_t^2)^2} \sum_i (t - x_i)^2 = \frac{N}{\sigma_x^2 +\sigma_t^2}$$

$$\sigma_t^2 = \frac{1}{N} \sum_i (t - x_i)^2 - \sigma_x^2$$

where $N$ is the number of observations of $X$.

The result for $\sigma_t$ says that the variance of the latent population is the sample variance minus the noise variance.

### $\sigma_i \ll \sigma_t$

In this limit the noise is negligible - each measurement very precisely measures the latent value, $t$. Then

$$t = \sum_i x_i$$

$$\sigma_t^2 = \frac{1}{N} \sum_i (t - x_i)^2$$,

i.e. the variance of the latent variable, $t$ is equal to the sample variance.

## multi-dimensional version

In more than 1-dimension, $t$ becomes a vector and $\sigma_i$ and $\sigma_t$ covariance matrices. The log-likelihood is then,

$$l = - 2 \ln{L} = \sum_i (x_i - t) (\Gamma_i + \Gamma_t)^{-1} (x_i - t)^{T} + \sum_i \ln{|\Gamma_i + \Gamma_t|}$$

where $\Gamma$ represents a covariance matrix and $|M|$ the determinant of the matrix $M$.

## numerical example - 1-dimension

Let’s say we have a population with mean $\mu=1$ and standard deviation $\sigma_t = 2$.

set.seed(101)
N = 10000
mu = 1
sigma_t = 2
population = rnorm(N, mu, sigma_t)


The noise distribution for each datum is drawn from a gamma distribution

noise_sd_shape = 20
noise_sd_rate = 10
noise_sd = rgamma(N, shape = noise_sd_shape, rate = noise_sd_rate)
data.frame(noise_sd=noise_sd) %>%
ggplot(aes(x=noise_sd)) +
geom_histogram() Now we can generate a synthetic set of measurements, applying this noise level.

noise = sapply(1:N, function(i) {rnorm(1, 0, noise_sd[i])})
measurements = population + noise


This sets up a log-likelihood function

log_likelihood_factory = function(measurements, noise_sd) {

function(params) {
t = params
sigma_t = params
logl_x = sum((measurements - t)**2 / (noise_sd ** 2 + sigma_t ** 2))
logl_c = sum(log(noise_sd ** 2 + sigma_t ** 2))
logl_x + logl_c
}
}


This finds the maximum likelihood values

logl_fun = log_likelihood_factory(measurements, noise_sd)
max_likelihood_solution = optim(c(1,1), logl_fun)
max_likelihood_solution$par #>  1.038575 1.953789  On the other hand if we look at the standard deviation of the measurements we get, sd(measurements) #>  2.840141  If we take the mean of the noise standard deviation as an effective value and apply the result for the case where all $\sigma_i$ are equal we get sqrt(var(measurements) - mean(noise_sd**2)) #>  1.969332  which is not a bad approximation. ## numerical example - 2-dimensions The following defines the parameters for a 2-dimensional latent population variable, with a correlation of $\rho = 0.5$ N = 10000 mu1 = 1 mu2 = 2 s1 = 1 s2 = 2 rho = 0.5  To simulate a correlated 2-d random variable we generate an uncorrelated variable and then multiply by the right-factor of a Cholesky decomposition. set.seed(101) D = 2 population_uncorr = matrix(rnorm(N * D, 0, 1), ncol=D) cor(population_uncorr) #> [,1] [,2] #> [1,] 1.000000000 0.005650561 #> [2,] 0.005650561 1.000000000  cov_mat = matrix(c(s1*s1, rho*s1*s2, rho*s1*s2, s2*s2), ncol=2) right_chol = chol(cov_mat) population = population_uncorr %*% right_chol population[,1] = population[,1] + mu1 population[,2] = population[,2] + mu2 cor(population) #> [,1] [,2] #> [1,] 1.0000000 0.5036932 #> [2,] 0.5036932 1.0000000  The noise distribution for each datum is drawn from uncorrelated gamma distributions noise_sd_shape = 20 noise_sd_rate = 10 noise_sd = matrix(rgamma(N * D, shape = noise_sd_shape, rate = noise_sd_rate), ncol=2)  Now we can generate a synthetic set of measurements, applying this noise level. noise = cbind( sapply(1:N, function(i) {rnorm(1, 0, noise_sd[i,1])}), sapply(1:N, function(i) {rnorm(1, 0, noise_sd[i,2])}) ) measurements = t( t(population) + t(noise))  We see that the noise applied to the population weakens the correlation df = data.frame(x1=measurements[,1], x2=measurements[,2], t1 = population[,1], t2=population[,2]) df %>% ggplot(aes(x=t1, y=t2)) + geom_point() + geom_point(aes(x=x1, y=x2), color='orange1', alpha=0.2) This sets up a log-likelihood function log_likelihood_factory_2d = function(measurements, noise_sd) { function(params) { tvec = matrix(c(params, params), ncol=2) s1 = params s2 = params rho = params cov_t = matrix(c(s1*s1, rho * s1*s2, rho*s1*s2, s2*s2), ncol=2) tmp = lapply(1:nrow(measurements), function(i) { cov_i = matrix(c(noise_sd[i,1]**2, 0, 0, noise_sd[i,2]**2), ncol=2) cov_mat = cov_i + cov_t right_chol = chol(cov_mat) right_chol_inv = solve(right_chol) w = matrix(measurements[i,] - tvec, nrow=1) %*% right_chol_inv logl_x = crossprod(t(w)) logl_c = determinant(cov_mat, logarithm = T)$modulus
list(logl_x=logl_x, logl_c = logl_c)
}) %>% bind_rows(

)

sum(tmp)

}
}


This finds the maximum likelihood values. I use the “L-BFGS-B” optimization method so I can explicitly put bounds on $\rho$.

logl_fun = log_likelihood_factory_2d(measurements, noise_sd)
max_likelihood_solution = optim(c(1, 1, 1, 1 , 0.5), logl_fun,
method = "L-BFGS-B",
lower = c(-Inf, -Inf, -Inf, -Inf, -1),
upper = c(Inf, Inf, Inf, Inf, 1)
)
max_likelihood_solution\$par
#>  1.0128679 2.0263088 0.9756733 1.9411863 0.5152542


The maximum likelihood method recovers the latent correlation of the population. On the other hand if we subtract a representative value for the data noise from the measurement sample covariance, we get a good approximation, as in the 1d case.

v1_x = var(noise[,1])
v2_x = var(noise[,2])
pop_covar_estimate = var(measurements) - matrix(c(v1_x, 0, 0, v2_x), ncol=2)
pop_covar_estimate
#>           [,1]      [,2]
#> [1,] 0.9472655 0.9734004
#> [2,] 0.9734004 3.8450709