Groupe 03 Projet DANJOU - DUROUSSEAU
+ +Exercise 1 : Negative weighted mixture
+Definition
+Question 1
+The conditions for a function f to be a probability density are :
+-
+
- f is defined on \(\mathbb{R}\) +
- f is non-negative, ie \(f(x) \ge +0\), \(\forall x \in +\mathbb{R}\) +
- f is Lebesgue-integrable +
- and \(\int_{\mathbb{R}} f(x) \,dx = +1\) +
The function f, to be a density, needs to be non-negative, ie \(f(|x|) \ge 0\) when \(|x| \rightarrow \infty\)
+We have, when \(|x| \rightarrow +\infty\), \(f_i(|x|) \sim +exp(\frac{-x^2}{\sigma_i^2})\) for \(i += 1, 2\)
+Then, \(f(|x|) \sim +exp(\frac{-x^2}{\sigma_1^2}) - exp(\frac{-x^2}{\sigma_2^2})\)
+We want \(f(|x|) \ge 0\), ie, \(exp(\frac{-x^2}{\sigma_1^2}) - +exp(\frac{-x^2}{\sigma_2^2}) \ge 0\)
+\(\Leftrightarrow \sigma_1^2 \ge +\sigma_2^2\)
+We can see that \(f_1\) dominates +the tail behavior.
+
+
Question 2
+For given parameters \((\mu_1, +\sigma_1^2)\) and \((\mu_2, +\sigma_2^2)\), we have \(\forall x \in +\mathbb{R}\), \(f(x) \ge 0\)
+\(\Leftrightarrow \frac{1}{\sigma_1} +exp(\frac{-(x-\mu_1)^2}{2 \sigma_1^2}) \ge \frac{a}{\sigma_2} +exp(\frac{-(x-\mu_2)^2}{2 \sigma_2^2})\)
+\(\Leftrightarrow 0 < a \le a^* = +\min_{x \in \mathbb{R}} \frac{f_1(x)}{f_2(x)} = \min_{x \in \mathbb{R}} +\frac{\sigma_2}{\sigma_1} exp(\frac{(x-\mu_2)^2}{2 \sigma_2^2} - +\frac{(x-\mu_1)^2}{2 \sigma_1^2})\)
+To find \(a^*\), we just have to +minimize \(g(x) := \frac{(x-\mu_2)^2}{2 +\sigma_2^2} - \frac{(x-\mu_1)^2}{2 \sigma_1^2}\)
+First we derive \(g\): \(\forall x \in \mathbb{R}, g^{\prime}(x) = \frac{x +- \mu_2}{\sigma_2^2} - \frac{x - \mu_1}{\sigma_1^2}\)
+We search \(x^*\) such that \(g^{\prime}(x^*) = 0\)
+\(\Leftrightarrow x^* = \frac{\mu_2 +\sigma_1^2 - \mu_1 \sigma_2^2}{\sigma_1^2 - \sigma_2^2}\)
+Then, we compute \(a^* = +\frac{f_1(x^*)}{f_2(x^*)}\)
+We call \(C \in \mathbb{R}\) the +normalization constant such that \(f(x) = C +(f_1(x) - af_2(x))\)
+To find \(C\), we know that \(1 = \int_{\mathbb{R}} f(x) \,dx = +\int_{\mathbb{R}} C (f_1(x) - af_2(x)) \,dx = C \int_{\mathbb{R}} f_1(x) +\,dx - Ca \int_{\mathbb{R}} f_2(x) \,dx = C(1-a)\) as \(f_1\) and \(f_2\) are density functions and by +linearity of the integrals.
+\(\Leftrightarrow C = +\frac{1}{1-a}\)
+
+
Question 3
+f <- function(a, mu1, mu2, s1, s2, x) {
+ fx <- dnorm(x, mu1, s1) - a * dnorm(x, mu2, s2)
+ fx[fx < 0] <- 0
+ return(fx / (1 - a))
+}
+
+a_star <- function(mu1, mu2, s1, s2) {
+ x_star <- (mu2 * s1^2 - mu1 * s2^2) / (s1^2 - s2^2)
+ return(dnorm(x_star, mu1, s1) / dnorm(x_star, mu2, s2))
+}mu1 <- 0
+mu2 <- 1
+s1 <- 3
+s2 <- 1
+
+x <- seq(-10, 10, length.out = 1000)
+as <- a_star(mu1, mu2, s1, s2)
+a_values <- c(0.1, 0.2, 0.3, 0.4, 0.5, as)
+
+plot(x, f(as, mu1, mu2, s1, s2, x),
+ type = "l",
+ col = "red",
+ xlab = "x",
+ ylab = "f(a, mu1, mu2, s1, s2, x)",
+ main = "Density function of f(a, mu1, mu2, s1, s2, x) for different a"
+)
+for (i in (length(a_values) - 1):1) {
+ lines(x, f(a_values[i], mu1, mu2, s1, s2, x), lty = 3, col = "blue")
+}
+legend("topright", legend = c("a = a*", "a != a*"), col = c("red", "blue"), lty = 1)
We observe that for small values of a, the density f is close to the +density of \(f_1 \sim \mathcal{N}(\mu_1, +\sigma_1^2)\). When a increases, the shape of evolves into the +combinaison of two normal distributions. We observe that for \(a = a^*\), the density the largest value of +a for which the density is still a density function, indeed for \(a > a^*\), the function f takes negative +values so it is no longer a density.
+s2_values <- seq(1, 10, length.out = 5)
+a <- 0.2
+
+plot(x, f(a, mu1, mu2, s1, max(s2_values), x),
+ type = "l",
+ xlab = "x",
+ ylab = "f(a, mu1, mu2, s1, s2, x)",
+ col = "red",
+ main = "Density function of f(a, mu1, mu2, s1, s2, x) for different s2"
+)
+
+for (i in length(s2_values):1) {
+ lines(x, f(a, mu1, mu2, s1, s2_values[i], x), lty = 1, col = rainbow(length(s2_values))[i])
+}
+
+legend("topright", legend = paste("s2 =", s2_values), col = rainbow(length(s2_values)), lty = 1)
We observe that when \(\sigma_2^2 = +1\), the density \(f\) has two +peaks and when \(\sigma_2^1 > 1\), +the density \(f\) has only one +peak.
+mu1 <- 0
+mu2 <- 1
+sigma1 <- 3
+sigma2 <- 1
+a <- 0.2
+as <- a_star(mu1, mu2, sigma1, sigma2)
+
+cat(sprintf("a* = %f, a = %f, a <= a* [%s]", as, a, a <= as))## a* = 0.313138, a = 0.200000, a <= a* [TRUE]We have \(\sigma_1^2 \ge +\sigma_2^2\) and \(0 < a \le +a^*\), so the numerical values are compatible with the +constraints defined above.
+Inverse c.d.f Random Variable simulation
+Question 4
+To prove that the cumulative density function F associated with f is +available in closed form, we need to compute \(F(x) = \int_{-\infty}^{x} f(t) \,dt = +\frac{1}{1-a} (\int_{-\infty}^{x} f_1(t)\, dt - a \int_{-\infty}^{x} +f_2(t)\, dt) = \frac{1}{1-a} (F_1(x) - aF_2(x))\) where \(F_1\) and \(F_2\) are the cumulative density functions +of \(f_1 \sim \mathcal{N}(\mu1, +\sigma_1^2)\) and \(f_2 \sim +\mathcal{N}(\mu2, \sigma_2^2)\) respectively.
+Then, \(F\) is a closed-form as a +finite sum of closed forms.
+F <- function(a, mu1, mu2, s1, s2, x) {
+ Fx <- pnorm(x, mu1, s1) - a * pnorm(x, mu2, s2)
+ return(Fx / (1 - a))
+}To construct an algorithm that returns the value of the inverse +function method as a function of u ∈(0,1), of the parameters \(a, \mu_1, \mu_2, \sigma_1, \sigma_2, x\), +and of an approximation precision \(\epsilon\), we can use the bisection +method.
+We fixe \(\epsilon > 0\).
+We set \(u \in (0, 1)\).
+We define \(L = -10\) and \(U = -L\), the bounds and \(M = \frac{L + U}{2}\), the middle of our
+interval.
+While \(U - L > \epsilon\) :
-
+
- We compute \(F(M) = \frac{1}{1-a} (F_1(M) +- aF_2(M))\) +
- If \(F(M) < u\), we set \(L = M\) +
- Else, we set \(U = M\) +
- We set \(M = \frac{L + U}{2}\) +
End while
+For the generation of random variables from F, we can use the inverse
+transform sampling method.
+
We set \(X\) as an empty array and
+\(n\) the number of random variables we
+want to generate.
+We fixe \(\epsilon > 0\).
+For \(i = 1, \dots, n\) :
-
+
We set \(u \in (0, 1)\).
+
+We define \(L = -10\) and \(U = -L\), the bounds and \(M = \frac{L + U}{2}\), the middle of our +interval.
+
+While \(U - L > \epsilon\) +:
+We compute \(F(M) = \frac{1}{1-a} +(F_1(M) - aF_2(M))\)
+If \(F(M) < u\), we set \(L = M\)
+Else, we set \(U = M\)
+We set \(M = \frac{L + +U}{2}\)
+End while
+We add \(M\) to \(X\)
+
End for We return \(X\)
+Question 5
+inv_cdf <- function(n) {
+ X <- numeric(n)
+ for (i in 1:n) {
+ u <- runif(1)
+ L <- -10
+ U <- -L
+ M <- (L + U) / 2
+ while (U - L > 1e-6) {
+ FM <- F(a, mu1, mu2, s1, s2, M)
+ if (FM < u) {
+ L <- M
+ } else {
+ U <- M
+ }
+ M <- (L + U) / 2
+ }
+ X[i] <- M
+ }
+ return(X)
+}set.seed(123)
+n <- 10000
+X <- inv_cdf(n)
+x <- seq(-10, 10, length.out = 1000)
+
+hist(X, breaks = 100, freq = FALSE, col = "lightblue", main = "Empirical density function", xlab = "x")
+lines(x, f(a, mu1, mu2, s1, s2, x), col = "red")
+legend("topright", legend = c("Theorical density", "Empirical density"), col = c("red", "lightblue"), lty = 1)
plot(ecdf(X), col = "blue", main = "Empirical cumulative distribution function", xlab = "x", ylab = "F(x)")
+lines(x, F(a, mu1, mu2, s1, s2, x), col = "red")
+legend("bottomright", legend = c("Theoretical cdf", "Empirical cdf"), col = c("red", "blue"), lty = 1)
We can see of both graphs that the empirical cumulative distribution +function is close to the theoretical one.
+Accept-Reject Random Variable simulation
+Question 6
+To simulate under f using the accept-reject algorithm, we need to
+find a density function g such that \(f(x) \le
+M g(x)\) for all \(x \in
+\mathbb{R}\), where M is a constant.
+
Then, we generate \(X \sim g\) and
+\(U \sim \mathcal{U}([0,1])\).
+We accept \(Y = X\) if \(U \le \frac{f(X)}{Mg(X)}\). Return \(1\) otherwise.
The probability of acceptance is \(\int_{\mathbb{R}} \frac{f(x)}{Mg(x)} g(x) \,dx = +\frac{1}{M} \int_{\mathbb{R}} f(x) \,dx = \frac{1}{M}\)
+Here we pose \(g = f1\).
+Then we have \(\frac{f(x)}{g(x)} = +\frac{1}{1-a} (1 - a\frac{f_2(x)}{f_1(x)})\)
+We pose earlier that \(a^* = \min_{x \in +\mathbb{R}} \frac{f_1(x)}{f_2(x)} \Rightarrow \frac{1}{a^*} = \max_{x +\in \mathbb{R}} \frac{f_2(x)}{f_1(x)}\).
+We compute in our fist equation : \(\frac{1}{1-a} (1 - a\frac{f_2(x)}{f_1(x)}) +\leq \frac{1}{1-a} (1 - \frac{a}{a^*}) \leq \frac{1}{1-a}\) +because \(a \leq a^* \Rightarrow 1 - +\frac{a}{a^*} \leq 0\)
+To conclude, we have \(M = +\frac{1}{1-a}\) and the probability of acceptance is \(\frac{1}{M} = 1-a\)
+Question 7
+accept_reject <- function(n, a) {
+ X <- numeric(0)
+ M <- 1 / (1 - a)
+
+ while (length(X) < n) {
+ Y <- rnorm(1, mu1, s1)
+ U <- runif(1)
+
+ if (U <= f(a, mu1, mu2, s1, s2, Y) / (M * dnorm(Y, mu1, s1))) {
+ X <- append(X, Y)
+ }
+ }
+ return(X)
+}set.seed(123)
+n <- 10000
+a <- 0.2
+X <- accept_reject(n, a)
+x <- seq(-10, 10, length.out = 1000)
+
+hist(X, breaks = 100, freq = FALSE, col = "lightblue", main = "Empirical density function", xlab = "x")
+lines(x, f(a, mu1, mu2, s1, s2, x), col = "red")
+legend("topright", legend = c("Theorical density", "Empirical density"), col = c("red", "lightblue"), lty = 1)
set.seed(123)
+
+acceptance_rate <- function(n, a = 0.2) {
+ Y <- rnorm(n, mu1, s1)
+ U <- runif(n)
+ return(mean(U <= f(a, mu1, mu2, s1, s2, Y) / (M * dnorm(Y, mu1, s1))))
+}
+
+M <- 1 / (1 - a)
+n <- 10000
+cat(sprintf("[M = %.2f] Empirical acceptance rate: %f, Theoretical acceptance rate: %f \n", M, acceptance_rate(n), 1 / M))## [M = 1.25] Empirical acceptance rate: 0.802600, Theoretical acceptance rate: 0.800000Question 8
+set.seed(123)
+a_values <- seq(0.01, 1, length.out = 100)
+acceptance_rates <- numeric(length(a_values))
+as <- a_star(mu1, mu2, s1, s2)
+
+for (i in seq_along(a_values)) {
+ acceptance_rates[i] <- acceptance_rate(n, a_values[i])
+}
+
+plot(a_values, acceptance_rates, type = "l", col = "blue", xlab = "a", ylab = "Acceptance rate", main = "Acceptance rate as a function of a")
+points(as, acceptance_rate(n, as), col = "red", pch = 16)
+legend("topright", legend = c("Acceptance rate for all a", "Acceptance rate for a_star"), col = c("lightblue", "red"), lty = c(1, NA), pch = c(NA, 16))