# Tag Archives: Machine learning

## Impact of correlated predictions on the variance of an ensemble model

Let $X_1$ and $X_2$ be the prediction errors of two statistical/machine learning algorithms. $X_1$ and $X_2$ have relatively low bias, and high variances $\sigma^2_1$ and $\sigma^2_2$. They are also correlated, having a Pearson correlation coefficient equal to $\rho_{1, 2}$.

Aggregating models $1$ and $2$ might result in a predictive model with lower prediction error variance than $1$ and $2$. But not all the times. For those who attended statistics/probability/portfolio optimization classes, this may probably sound obvious; you can directly jump to the illustrative R part, below.

Let $Z := \alpha X_1 + (1-\alpha) X_2$, with $\alpha \in \left[0, 1\right]$, be the prediction error of the ensemble model built with $1$ and $2$. We have :

$Var(Z) = \alpha^2 \sigma^2_1 + (1 - \alpha)^2\sigma^2_2 + 2\alpha(1-\alpha)Cov(X_1, X_2)$

And from the fact that :

$Cov(X_1, X_2) = \rho_{1, 2} \sigma_1\sigma_2$

We get :

$Var(Z) = \alpha^2 \sigma^2_1 + (1 - \alpha)^2\sigma^2_2 + 2\alpha(1-\alpha) \rho_{1, 2} \sigma_1\sigma_2$

Now, let’s see how $Var(Z)$ changes with an increase of $\alpha$, the ensemble’s allocation for model $1$ :

$\frac{\partial Var(Z)}{\partial \alpha}= 2\alpha \sigma^2_1 - 2 (1 - \alpha) \sigma^2_2 + 2(1-2\alpha) \rho_{1, 2} \sigma_1\sigma_2$

When $\alpha$ is close to $0$, that is, when the ensemble contains almost only model $2$, we have :

$\frac{\partial Var(Z)}{\partial \alpha}_{|\alpha = 0}= 2 \left( \rho_{1, 2} \sigma_1\sigma_2 - \sigma^2_2 \right) = 2 \sigma^2_2\left( \rho_{1, 2} \frac{\sigma_1}{\sigma_2} - 1 \right)$

That’s the relative change in the variance of the ensemble prediction error, induced by introducing model $1$ in an ensemble containing almost only $2$. Hence, if $\rho_{1, 2} = \frac{\sigma_2}{\sigma_1}$, increasing the allocation of model $1$ won’t increase or decrease the variance of ensemble prediction at all. The variance will decrease if we introduce in the ensemble a model $1$, so that $\rho_{1, 2} \leq \frac{\sigma_2}{\sigma_1}$. If $\rho_{1, 2} \geq \frac{\sigma_2}{\sigma_1}$, it won’t decrease, no matter the combination of $X_1$ and $X_2$.

For a simple illustrative example in R, I create simulated data observations :

  # number of observations
n <- 100
u <- 1:n

# Simulated observed data
intercept <- 5
slope <- 0.2
set.seed(2)

## data
y <-  intercept + slope*u + rnorm(n)
plot(u, y, type = 'l', main = "Simulated data observations")
points(u, y, pch = 19)


I fit a linear regression model to the data, as a benchmark model :

  # Defining training and test sets
train <- 1:(0.7*n)
test <- -train
u.train <- u[(train)]
y.train <- y[(train)]
u.test <- u[(test)]
y.test <- y[(test)]

# Fitting the benchmark model to the training set
fit.lm <- lm(y.train ~ u.train)

(slope.coeff <- fit.lm$coefficients[2]) (intercept.coeff <- fit.lm$coefficients[1])

## u.train
## 0.1925
## (Intercept)
## 5.292


Obtaining the predicted values from the benchmark model, and prediction error

  # Predicted values from benchmark model on the test set
y.pred <- intercept.coeff + slope.coeff*u.test

# prediction error from linear regression
pred.err <- y.pred - y.test
(mean.pred.err <- mean(pred.err))
(var.pred.err <- var(pred.err))

## [1] -0.1802
## [1] 1.338


Now, I consider two other models, $1$ and $2$ :

$y = a + b \times u + sin(u) + \epsilon_1$

and

$y = a + b \times u - 0.35 \times sin(u) + \epsilon_2$

with $\epsilon_1$ and $\epsilon_2$ being 2 correlated gaussians with zero mean and constant variances (well, not really “models” that i fit, but these help me to build fictitious various predictions with different correlations for the illustration). The slope and intercept are those obtained from the benchmark model.

 # Alternative model 1 (low bias, high variance, oscillating)
m <- length(y.pred)
eps1 <- rnorm(m, mean = 0, sd = 1.5)
y.pred1 <- intercept.coeff + slope.coeff*u.test + sin(u.test) + eps1

# prediction error for model 1
pred.err1 <- y.pred1 - y.test


We can visualize the predictions of model $1$, $2$ and the benchmark, with different prediction errors correlations between model $1$ and $2$, to get an intuition of the possible ensemble predictions :

  # Different prediction errors correlations for 1 and 2
rho.vec <- c(-1, -0.8, 0.6, 1)

# Independent random gaussian numbers defining model 2 errors
eps <- rnorm(m, mean = 0, sd = 2)

# Plotting the predictions with different correlations
par(mfrow=c(2, 2))
for (i in 1:4)
{
rho <- rho.vec[i]

# Correlated gaussian numbers (Cholesky decomposition)
eps2 <- rho*eps1 + sqrt(1 - rho^2)*eps

# Alternative  model 2 (low bias, higher variance than 1, oscillating)
y.pred2 <- intercept.coeff + slope.coeff*u.test - 0.35*sin(u.test) + eps2

# prediction error for model 2
pred.err2 <- y.pred2 - y.test

# predictions from 1 & 2 correlation
corr.pred12 <- round(cor(pred.err1, pred.err2), 2)

# Plot
plot(u.test, y.test, type = "p",
xlab = "test set values", ylab = "predicted values",
main = paste("models 1 & 2 pred. values \n correlation :",
corr.pred12))
points(u.test, y.test, pch = 19)
lines(u.test, y.pred, lwd = 2)
lines(u.test, y.pred1,
col = "blue", lwd = 2)
lines(u.test, y.pred2,
col = "red", lwd = 2)
}


Now including allocations of models $1$ and $2$, we can see how the ensemble variance evolves as a function of allocation and correlation :

  # Allocation for model 1 in the ensemble
alpha.vec <- seq(from = 0, to = 1, by = 0.05)

# Correlations between model 1 and model 2
rho.vec <- seq(from = -1, to = 1, by = 0.05)

# Results matrices
nb.row <- length(alpha.vec)
nb.col <- length(rho.vec)

## Average prediction errors of the ensemble
mean.pred.err.ens <- matrix(0, nrow = nb.row, ncol = nb.col)
rownames(mean.pred.err.ens) <- paste0("pct. 1 : ", alpha.vec*100, "%")
colnames(mean.pred.err.ens) <- paste0("corr(1, 2) : ", rho.vec)

## Variance of prediction error of the ensemble
var.pred.err.ens <- matrix(0, nrow = nb.row, ncol = nb.col)
rownames(var.pred.err.ens) <- paste0("pct. 1 : ", alpha.vec*100, "%")
colnames(var.pred.err.ens) <- paste0("corr(1, 2) : ", rho.vec)

# loop on correlations and allocations
for (i in 1:nb.row)
{
for (j in 1:nb.col)
{
alpha <- alpha.vec[i]
rho <- rho.vec[j]

# Alternative model 2 (low bias, higher variance, oscillating)
eps2 <- rho*eps1 + sqrt(1 - rho^2)*eps
y.pred2 <- intercept.coeff + slope.coeff*u.test - 0.35*sin(u.test) + eps2
pred.err2 <- y.pred2 - y.test

# Ensemble prediction error
z <- alpha*pred.err1 + (1-alpha)*pred.err2
mean.pred.err.ens[i, j] <- mean(z)
var.pred.err.ens[i, j] <-  var(z)
}
}
res.var <- var.pred.err.ens

# Heat map for the variance of the ensemble
filled.contour(alpha.vec, rho.vec, res.var, color = terrain.colors,
main = "prediction error variance for the ensemble",
xlab = "allocation in x1", ylab = "correlation between 1 and 2")


Hence, the lower the correlation between $1$ and $2$, the lower the variance of the ensemble model prediction. This, combined with an allocation of $1$ comprised between $35\%$ and $50\%$ seem to be the building blocks for the final model. The ensemble models biases helps in making a choice of allocation (this is actually an optimization problem that can be directly solved with portfolio theory).

Finding the final ensemble, with lower variance and lower bias :

# Final ensemble

## Allocation
alpha.vec[which.min(as.vector(res.var))]
## [1] 0.45

## Correlation
res.bias <- abs(mean.pred.err.ens)
which.min(res.bias[which.min(as.vector(res.var)), ])
## corr(1, 2) :
-0.7
## 7


Creating the final model with these parameters :

    rho <- -0.7
eps2 <- rho*eps1 + sqrt(1 - rho^2)*eps

# Alternative model 2 (low bias, higher variance, oscillating)
y.pred2 <- intercept.coeff + slope.coeff*u.test - 0.35*sin(u.test) + eps2

# Final ensemble prediction
y.pred.ens <- 0.45*y.pred1 + 0.55*y.pred2

# Plot
plot(u.test, y.test, type = "p",
xlab = "test set", ylab = "predicted values",
main = "Final ensemble model (green)")
points(u.test, y.test, pch = 19)
# benchmark
lines(u.test, y.pred, lwd = 2)
# model 1
lines(u.test, y.pred1, col = "blue", lwd = 2)
# model 2
lines(u.test, y.pred2, col = "red", lwd = 2)
# ensemble model with 1 and 2
points(u.test, y.pred.ens, col = "green", pch = 19)
lines(u.test, y.pred.ens, col = "green", lwd = 2)


Performance of the final model :

    # Benchmark
pred.err <- y.pred - y.test
# Model 1
pred1.err <- y.pred1 - y.test
# Model 2
pred2.err <- y.pred2 - y.test
# Ensemble model
pred.ens.err <- y.pred.ens - y.test

# Bias comparison
bias.ens <- mean(y.pred.ens - y.test)
bias.ens_vsbench <- (bias.ens/mean(y.pred - y.test) - 1)*100
bias.ens_vs1 <- (bias.ens/mean(y.pred1 - y.test) - 1)*100
bias.ens_vs2 <- (bias.ens/mean(y.pred2 - y.test) - 1)*100

# Variance comparison
var.ens <- var(y.pred.ens - y.test)
var.ens_vsbench <- (var.ens/var(y.pred - y.test) - 1)*100
var.ens_vs1 <- (var.ens/var(y.pred1 - y.test) - 1)*100
var.ens_vs2 <- (var.ens/var(y.pred2 - y.test) - 1)*100

cbind(c(bias.ens_vsbench, bias.ens_vs1, bias.ens_vs2), c(var.ens_vsbench, var.ens_vs1, var.ens_vs2))
##        [,1]  [,2]
## [1,] -95.31 46.27
## [2,] -112.12 -62.93
## [3,] -88.33 -45.53


Filed under R stuff

## Solvency II yield curve ‘fits exactly’. Too tightly to explain ?

For the QIS5 and the LTGA (Quantitative Impact Studies), the Prudential Authority (EIOPA) suggested the use of Smith-Wilson method for the purpose of discounting future cash flows. A description of this method can be found here. The technical specifications for the QIS5 argued about this method that : “the term structure is fitted exactly to all observed zero coupon bond prices”. The method was subsequently used for the LTGA.

But : is fitting exactly, a desirable feature for a yield curve ? I won’t give a definitive answer here, but some hints, as the question could be answered in a lot of different ways (well… this post is mainly an excuse to present you ycinterextra in action after this, for you to tell me how to improve the package).

I’ll fit the Nelson-Siegel (NS), Svensson (SV) and Smith-Wilson (SW) models (implemented in ycinterextra) to the data from Diebold, F. X., & Li, C. (2006). Forecasting the term structure of government bond yields. Journal of econometrics, 130(2), 337-364. And  we’ll see how accurate are the  values they produce on out-of-sample data. I used mixed ideas from Diebold and Li (2006), along with Gilli, Manfred and Schumann, Enrico, A Note on ‘Good Starting Values’ in Numerical Optimisation (June 3, 2010). Available at SSRN: http://ssrn.com/abstract=1620083.

The data can be found here : http://www.ssc.upenn.edu/~fdiebold/papers/paper49/FBFITTED.txt.

# I put the data in a text file.
data.diebold.li <- scan("FBFITTED.txt", skip = 14)


library(ycinterextra)

# Number of observation dates from January 1985 through December 2000
nbdates <- 372

# Time to maturities for each observation date
mats <- round(c(1, 3, 6, 9, 12, 15, 18, 21, 24, 30, 36, 48, 60, 72, 84, 96, 108, 120)/12, 3)
nbmats <- length(mats)
nbcol <- nbmats + 1

# Formatting the data
data.diebold.li.final <- NULL
for(i in seq_len(nbdates))
{
data.diebold.li.final <- rbind(data.diebold.li.final, data.diebold.li[(nbcol*(i -1)+1):(nbcol*(i -1)+nbcol)])
}


A 3D plot of the observed data can be obtained with the following function :

x <- mats
y <- data.diebold.li.final[,1]
z <- data.diebold.li.final[,-1]

graph3D <- function(x, y, z)
{
par(bg = "white")
nrz <- nrow(z)
ncz <- ncol(z)
# Create a function interpolating colors in the range of specified colors
jet.colors <- colorRampPalette( c("blue", "green") )
# Generate the desired number of colors from this palette

nbcol <- 100
color <- jet.colors(nbcol)
# Compute the z-value at the facet centres
zfacet <- z[-1, -1] + z[-1, -ncz] + z[-nrz, -1] + z[-nrz, -ncz]
# Recode facet z-values into color indices
facetcol <- cut(zfacet, nbcol)
persp(y, x, z, col = color[facetcol], phi = 30, theta = -30, ticktype= 'detailed', nticks = 3,
xlab = "observation date", ylab = "maturity", zlab = "zero rates in pct.")
}

graph3D(x, y, z)


Now, let’s calibrate the 3 models, NS, SV and SW to the observed yield curves (this takes a lot of time, as there are 372 curves, each with 18 maturities) :

nrz <- nrow(z)
ncz <- ncol(z)
yM.NS <- matrix(NA, nrow = nrz, ncol = ncz)
yM.SV <- matrix(NA, nrow = nrz, ncol = ncz)
yM.SW <- matrix(NA, nrow = nrz, ncol = ncz)

# Loop over the dates from January 1985 through December 2000
for (t in seq_len(nbdates))
{
# market yields
yM <- as.numeric(data.diebold.li.final[t, -1])

yM.NS[t, ] <- fitted(ycinter(yM = yM, matsin = mats, matsout = mats,
method = "NS", typeres = "rates"))

yM.SV[t, ] <- fitted(ycinter(yM = yM, matsin = mats, matsout = mats,
method = "SV", typeres = "rates"))

yM.SW[t, ] <- fitted(ycinter(yM = yM, matsin = mats, matsout = mats,
method = "SW", typeres = "rates"))
}


After a little time… SW curve seems to fit very well indeed, and the plot below looks pretty similar to the one we made before :

graph3D(x, y, yM.SW)


With a plot of the average yield curve (average of the 372 observed curves as in Diebold and Li (2006)), we can see that all of the three models seem to provide fairly good fits to the data :

# Average yield curves
plot(mats, apply(z, 2, mean), pch = 3, xlab = "time to maturity",
ylab = "yield to maturity",
main = paste("Actual and fitted (model-based)", "\n", "average yield curves"))
lines(mats, apply(yM.NS, 2, mean), col = "blue", lwd= 2)
lines(mats, apply(yM.SV, 2, mean), col = "red", lwd= 2)
lines(mats, apply(yM.SW, 2, mean), col = "green", lwd=2, lty = 2)
legend("bottomright", c("Actual", "NS", "SV", "SW"),
col=c("black", "blue", "red", "green"), text.col = c("black", "blue", "red", "green"),
lty=c(NA, 1, 1, 2), lwd = c(1, 2, 2, 2), pch=c(3, NA, NA, NA))


Here are plots of the residuals for NS and SW (for SV, the plot is pretty similar to the plot for NS) :

# residuals
graph3D(x, y, yM.NS-z)
graph3D(x, y, yM.SW-z)


Now, taking a closer look at some given cross-sections (here on 3/31/1989), we see what’s actually going on (click for a better resolution) :

t <- 231

par(mfrow=c(2,2))

plot(mats, data.diebold.li.final[t, -1], xlab = "time to maturity",
ylab = "yield to maturity", col="red", main = "Observed data")
points(mats, data.diebold.li.final[t, -1], col="red")

plot(mats, yM.NS [t,], type='l', xlab = "time to maturity",
ylab = "yield to maturity", col = "blue", main = "Nelson-Siegel fit")
points(mats, data.diebold.li.final[t, -1], col="red")

plot(mats, yM.SV[t,], type='l', xlab = "time to maturity",
ylab = "yield to maturity", col = "blue", main = "Svensson fit")
points(mats, data.diebold.li.final[t, -1], col="red")

plot(mats, yM.SW[t,], type='l', xlab = "time to maturity",
ylab = "yield to maturity", col = "blue", main = "Smith-Wilson fit")
points(mats, data.diebold.li.final[t, -1], col="red")



And for 7/31/1989 (with t == 235) :

NS fits, well. SV fits a litte bit better, due to its extended number of parameters. SW fits perfectly to each point, due to its laaarge number of parameters.

Let’s see how accurate are the values produced by the 3 methods on out-of-sample values, by the use of a simple machine learning test set method (a  method like k-fold cross-validation can be used as well, it’s computationnally more expensive, but simple though). I divide the sample data into 2 sets :

•  A learning set with the zero rates observed at time to maturities : c(1, 6, 9, 15, 18, 21, 24, 30, 48, 72, 84, 96, 120)/12
• A test set with the zero rates observed at time to maturities : c(3, 12, 36, 60, 108)/12

The models’ parameters are estimated using the learning set, and the test set is used to assess how well (with less error) the model can predict unknown values :

# Maturities for test set
matsoutsample <- c(3, 12, 36, 60, 108)/12
matchoutsample <- match(matsoutsample, mats)
m <- length(matsoutsample)

# Maturities for learning set
matsinsample <- mats[-matchoutsample]


# Residuals matrices
res.NS <- matrix(NA, nrow = nrz, ncol = length(matchoutsample))
res.SV <- matrix(NA, nrow = nrz, ncol = length(matchoutsample))
res.SW <- matrix(NA, nrow = nrz, ncol = length(matchoutsample))

# Loop over the dates from January 1985 through December 2000
for (t in seq_len(nbdates))
{
yM.t <- as.numeric(data.diebold.li.final[t, -c(1, matchoutsample)])

yM.t.bechm <- as.numeric(data.diebold.li.final[t, -1])

res.NS[t, ] <- (fitted(ycinter(yM = yM.t, matsin = matsinsample, matsout = mats,
method = "NS", typeres = "rates")) - yM.t.bechm)[matchoutsample]

res.SV[t, ] <- (fitted(ycinter(yM = yM.t, matsin = matsinsample, matsout = mats,
method = "SV", typeres = "rates")) - yM.t.bechm)[matchoutsample]

res.SW[t, ] <- (fitted(ycinter(yM = yM.t, matsin = matsinsample, matsout = mats,
method = "SW", typeres = "rates")) - yM.t.bechm)[matchoutsample]
}


For each model, we get the 372 mean squared errors as follows :

mse.ns <- apply((res.NS), 1, function(x) crossprod(x))/m
mse.sv <- apply((res.SV), 1, function(x) crossprod(x))/m
mse.sw <- apply((res.SW), 1, function(x) crossprod(x))/m


A summary of these mean squared errors shows that SW actually overfits, and fails to predict the out-of-sample data :

summary(mse.ns)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.0005895 0.0054050 0.0117700 0.0227100 0.0198400 0.5957000
summary(mse.sv)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.0003694 0.0075310 0.0154200 0.0329400 0.0319700 0.6079000
summary(mse.sw)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.057 0.556 1.478 431.200 10.820 29040.000