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I am having trouble in simulating values for a random variable $X$ having a piecewise gamma failure rate: $$ \lambda_X(t) =\lambda_1(t)1\!\!1_{\lbrace t\leq t_0 \rbrace} + \lambda_2(t)1\!\!1_{\lbrace t > t_0 \rbrace}$$ where $\lambda_1$ and $\lambda_2$ are the failure rates of two Gamma distributions $\Gamma(\alpha_1,\beta_1)$ and $\Gamma(\alpha_2,\beta_2)$ respectively. The value of $t_0$ and the parameters of Gamma distributions are known in advance.

Apparently the change point $t_0$ doesnot allow simulating two independant gamma random variables. Does anyone have a solution? Thanks in advance.

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  • $\begingroup$ Do you want to simulate a random variable X at time $t$ or a stochastic process $(X_t)$ in continuous time or do you want to invert the failure rate into a distribution ? $\endgroup$ – Xi'an Jun 25 '15 at 16:52
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    $\begingroup$ I want to simulate samples $x_1,x_2,...,x_n$ for the r.v. X. I think you hit the right spot when proposing "invert the failure rate into a distribution", because that what I am taught in simulation courses for simple distributions like the normal or exponential (then using the probability transformation property to simulate the samples). However, I am not sure how to proceed for this kind of distribution $\endgroup$ – Tuan Jun 25 '15 at 17:01
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As a general rule, when given the failure rate$$\eta(t)=\frac{f(t)}{\int_t^\infty f(x)\,\text{d}x}=-\frac{\text{d}}{\text{d}t}\,\log \int_t^\infty f(x)\,\text{d}x$$we can formally derive$$\int_0^t\eta(x)\,\text{d}x=-\log \int_t^\infty f(x)\,\text{d}x$$since the value at $t=0$ is zero on both sides, and$$\exp\left\{-\int_0^t\eta(x)\,\text{d}x\right\}=1-\int_0^t f(x)\,\text{d}x$$which leads to the cdf$$F(t)=\int_0^t f(x)\,\text{d}x=1-\exp\left\{-\int_0^t\eta(x)\,\text{d}x\right\}$$From a simulation perspective, if we generate $U\sim\mathcal{U}(0,1)$, we have to solve $F(X)=U$, which means $$1-\exp\left\{-\int_0^X\eta(x)\,\text{d}x\right\}=U$$or$$\int_0^X\eta(x)\,\text{d}x=-\log(1-U)$$ Now, when $\eta(x)=\lambda_1(x)\mathbb{I}_{x\le t_0}+\lambda_2(x)\mathbb{I}_{x>t_0}$, this means that $$F(x)=\cases{F_1(x) &\text{when }x<t_0\\ 1-\{1-F_1(t_0)\}\dfrac{1-F_2(x)}{1-F_2(t_0)} &\text{else}}$$ where $F_1$ is the Gamma $G(\alpha_1,\beta_1)$ cdf and $F_2$ the Gamma $G(\alpha_2,\beta_2)$ cdf. Equivalently, $$F(x)=\cases{F_1(t_0)\mathbb{P}(X_1\le x|X_1\le t_0) &\text{when }x<t_0\\ F_1(t_0)+\{1-F_1(t_0)\}\mathbb{P}(X_2\le x|X_2\ge t_0) &\text{else}}$$where $X_1$ and $X_2$ denote Gamma $G(\alpha_1,\beta_1)$ and Gamma $G(\alpha_2,\beta_2)$ random variables.

Conclusion This representation of the cdf establishes that the distribution with failure rate$$\eta(x)=\lambda_1(x)\mathbb{I}_{x\le t_0}+\lambda_2(x)\mathbb{I}_{x>t_0}$$is a mixture of a Gamma $G(\alpha_1,\beta_1)$ truncated to $(0,t_0)$ and of a Gamma $G(\alpha_2,\beta_2)$ truncated to $(t_0,\infty)$ with weights $F_1(t_0)$ and $(1-F_1(t_0))$.

In terms of random simulation this means

  1. pick the generating distribution Gamma $G(\alpha_1,\beta_1)$ truncated to $(0,t_0)$ with probability $F_1(t_0)$ and the generating distribution Gamma $G(\alpha_2,\beta_2)$ truncated to $(t_0,\infty)$ with probability $1-F_1(t_0)$
  2. Generate X from the picked truncated Gamma distribution
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  • $\begingroup$ If I understand correctly, you are suggesting that the r.v. $X$ is a mixture of two independant and weighted truncated Gamma distributions? Following your developpement, the question (and solution) can be generalized to almost any distribution. $\endgroup$ – Tuan Jun 25 '15 at 18:25
  • $\begingroup$ I am worried by your use of "independent": the density of the distribution is of the form$$F_1(t_0)g_1(x)+(1-F_1(t_0))g_2(x)$$where $g_1$ and $g_2$ are the densities of the two truncated gammas. There is no notion of "independence" involved. $\endgroup$ – Xi'an Jun 25 '15 at 19:30
  • $\begingroup$ Thanks for your concern. What I meant by independent (maybe I am wrong) is that it seems there is no connection between $g_1$ and $g_2$ (well there is $t_0$ but not in the sense of conditional probability for example). My intuition should be wrong, since your reasoning is flawfless. $\endgroup$ – Tuan Jun 25 '15 at 20:32

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