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Consider the following random variable $$ Z=\min_i\{X_i+Y_i\} $$ for $-n\leq i\leq n$, where $X_i\overset{\mathrm{iid}}{\sim}\text{Exp}(\lambda)$, $Y_i\overset{\mathrm{iid}}{\sim}\text{Erlang}(|i|,\gamma)$, and $X_i \perp Y_i,\forall i$.

One can think of $Z$ as the time of first arrival (at $i=0$) of incoming waves starting from different locations ($i$). For each $i$, we assume it takes time $X_i$ to initiate a wave at point $i$ and time $Y_i$ to for the wave to reach $i=0$. $Z$ is the minimum of all possible times.

What can we say about $\mathbb{E}[Z]$?

Some ideas: In the case where we take $Y_i$ as its expected value, that is, $\mathbb{E}[Y_i]=|i|/\gamma$, we may actually compute a closed formula for $\mathbb{E}[Z]$, assuming non-negative time, $$\tag{1} \mathbb{E}[Z]=\mathbb{E}[\min_i\{X_i+|i|/\gamma\}]=\int_0^\infty \prod_i \min \{ 1,\exp(-\lambda(t-|i|/\gamma)) \} \,dt $$ which can then be split into a series of different integrals, and where the product is taken over all $i$. For full derivation details, see this answer in the context of an activation process.

If this "wave arrival time", $Y_i$, follows an Erlang distribution, the problem becomes slightly more complex. However, averaged simulations of the activation process under these distribution assumption say otherwise (that is, expression (1) is a good enough approximation). So there must be not much difference between considering the pure random variables, or its expected value. I would like to show this mathematically, either by deducing a similar expression if I take an Erlang-distributed random variable, or by other means. Any ideas?

Further insights: From this paper, we have the following result

Corollary 6.2. Let $Z = Y_1 + \cdots + Y_k$ be the sum of $k$ independent random variables having the Erlang distribution with parameter $(m_j, \gamma_j)$ for $j = 1, \dots, k$, and $\gamma_j \neq \gamma_p$ for $j \neq p$. Then the density for $Z$, $f_Z(t)$, for $t \geq 0$ is given by $$ f_Z(t) = \prod_{j=1}^k \gamma_j^{m_j} \left\{ \sum_{j=1}^k e^{-\gamma_j t} (-1)^{m_j-1} \times \sum_{\sum_{p=1}^k r_p = m_j-1, \, r_p \geq 0} \frac{(-t)^{r_j}}{r_j!} \prod_{q=1, q \neq j}^k \frac{(m_q + r_q - 1)!}{(\gamma_q - \gamma_j)^{m_q + r_q}} \right\}. $$ In our case, let $Z_i\equiv X_i+Y_i$ and set $k=2$, $(m_1,\gamma_1)=(1,\lambda)$, and $(m_2,\gamma_2)=(|i|,\gamma)$, the CDF $F_{Z_i}(t)$ is given by $$ F_{Z_i}(t) = 1 - e^{-\lambda t} + \lambda \gamma^{|i|} \sum_{r_2=0}^{|i|-1} \frac{r_1!}{(\lambda - \gamma)^{1 + r_1}} \left( \frac{\Gamma(r_2+1,0) - \Gamma(r_2+1, \gamma t)}{\gamma^{r_2+1}} \right) $$ where $\Gamma$ is the incomplete Gamma function, given by $$ \Gamma(s, z) = \int_z^\infty x^{s-1} e^{-x} \, dx, \quad \text{for } s > 0 \text{ and } z \geq 0. $$ From this question, we know that if the CDF of $Z_i$ is denoted by $F_{Z_i}(t)$, then the CDF of the minimum $Z$ is given by $F_Z(t)=1-[1-F_{Z_i}(t)]^{2n+1}$. Hence, $$ \begin{align} \mathbb{E}[Z]&=\int_0^\infty (1-F_Z(t))\,dt\\ &= \int_0^\infty \left( e^{-\lambda t} - \lambda \gamma^{|i|} \sum_{r_2=0}^{|i|-1} \frac{r_1!}{(\lambda - \gamma)^{1 + r_1}} \left( \frac{\Gamma(r_2+1, 0) - \Gamma(r_2+1, \gamma t)}{\gamma^{r_2+1}} \right) \right)^{2n+1} \, dt. \end{align} $$ I wonder if there is a simplification somewhere.

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  • $\begingroup$ Have you looked at classical asymptotics of the min in the large n limit? The asymptotics are dictated by the shape of the tail of the pdf. $\endgroup$ Commented Oct 9, 2024 at 9:53
  • $\begingroup$ Is $Y_0$ always equal to zero? $\endgroup$ Commented Oct 10, 2024 at 4:29
  • $\begingroup$ The mean of $Z$ for $n=1$ seems to be $\frac{1}{9} \left(4 \left(\frac{1}{\gamma +2 \lambda }+\frac{1}{2 \gamma +\lambda }\right)+\frac{3}{\lambda }\right)$ if $\lambda \neq \gamma$. $\endgroup$ Commented Oct 11, 2024 at 6:22
  • $\begingroup$ The mean of $Z$ for $n=2$ seems to be $$\frac{13824 \gamma ^{11}+293760 \gamma ^{10} \lambda +2331360 \gamma ^9 \lambda ^2+9638920 \gamma ^8 \lambda ^3+23561620 \gamma ^7 \lambda ^4+36189166 \gamma ^6 \lambda ^5+35876495 \gamma ^5 \lambda ^6+23033010 \gamma ^4 \lambda ^7+9398420 \gamma ^3 \lambda ^8+2319480 \gamma ^2 \lambda ^9+311040 \gamma \lambda ^{10}+17280 \lambda ^{11}}{5 \lambda (4 \gamma +\lambda )^3 (3 \gamma +2 \lambda )^3 (2 \gamma +3 \lambda )^3 (\gamma +4 \lambda )^2}$$. $\endgroup$ Commented Oct 12, 2024 at 6:06
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    $\begingroup$ Symbolic means will likely get much more complicated with increasing values of $n$. But obtaining the means numerically for specific values of $n$, $\lambda$, and $\gamma$ appears doable. Are you still interested in this problem? $\endgroup$ Commented Oct 12, 2024 at 6:08

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If $X_i\sim \text{Exponential}(\lambda)$ and $Y_i \sim \text{Erlang}(|i|,\gamma)$ for $i \neq 0$ and $\gamma>\lambda$, then the pdf of $Z_i=X_i+Y_i$ is given by

$$f_i(z)=\int_0^z \lambda e^{-\lambda x} \times \frac{\gamma ^{|i|} (z-x)^{|i|-1} e^{\gamma (x-z)}}{\Gamma (|i|)} dx=\frac{\lambda \gamma ^{|i|} e^{\lambda (-z)} (\gamma -\lambda )^{-|i|} (\Gamma (|i|)-\Gamma (|i|,z (\gamma -\lambda )))}{\Gamma (|i|)}$$

We have $f_{i}(z)=f_{-i}(z)$. The cdf for $Z_i$ is then

$$F_i (z)=\int_0^z f_i(t)dt=\frac{e^{\lambda (-z)} \left(\frac{\gamma }{\gamma -\lambda }\right)^{|i|} (\Gamma (|i|,z (\gamma -\lambda ))-\Gamma (|i|))-e^{\gamma (-z)} (\gamma z)^{|i|-1}-(|i|-1) \Gamma (|i|-1,z \gamma )+\Gamma (|i|)}{\Gamma (|i|)}$$

Define $Z=\min_{-n\leq i \leq n}\{Z_i\}$. The cdf for $Z$ is

$$\text{Pr}(Z\leq z)=1-(1-F_0(z)) \prod _{i=1}^n (1-F_i(z))^2$$

and the pdf for $Z$ is the derivative of the cdf with respect to $z$.

Using Mathematica we have the following starting with the cdf's for $Z_i$:

F[0] = 1 - E^(- λ z);
F[i_] := (-E^(-z γ) (z γ)^(-1 + i) + Gamma[i] - (-1 + i) Gamma[-1 + i, z γ] + 
  E^(-z λ) (γ/(γ - λ))^i (-Gamma[i] + Gamma[i, z (γ - λ)]))/Gamma[i]

This results in functions for the cdf, pdf, and mean of $Z$:

cdf[n_] := 1 - (1 - F[0]) Product[(1 - F[i])^2, {i, 1, n}]
pdf[n_] := FullSimplify[D[cdf[n], z], Assumptions -> {n ∈ PositiveIntegers, γ > λ > 0, z > 0}]

mean[n_] := Integrate[z  pdf[n], {z, 0, ∞}, 
  Assumptions ->{n ∈ PositiveIntegers, γ > λ > 0, z > 0}]

mean[1]

$$\frac{1}{9} \left(4 \left(\frac{1}{\gamma +2 \lambda }+\frac{1}{2 \gamma +\lambda }\right)+\frac{3}{\lambda }\right)$$

mean[2]

$$\frac{13824 \gamma ^{11}+293760 \gamma ^{10} \lambda +2331360 \gamma ^9 \lambda ^2+9638920 \gamma ^8 \lambda ^3+23561620 \gamma ^7 \lambda ^4+36189166 \gamma ^6 \lambda ^5+35876495 \gamma ^5 \lambda ^6+23033010 \gamma ^4 \lambda ^7+9398420 \gamma ^3 \lambda ^8+2319480 \gamma ^2 \lambda ^9+311040 \gamma \lambda ^{10}+17280 \lambda ^{11}}{5 \lambda (4 \gamma +\lambda )^3 (3 \gamma +2 \lambda )^3 (2 \gamma +3 \lambda )^3 (\gamma +4 \lambda )^2}$$

For $n > 2$ the symbolic results are likely too complicated and lengthy to be useful. But placing numeric values for $\lambda$ and $\gamma$ can produce nearly exact results for values of $n > 2$. For example, consider $n = 5$ and $\lambda = 1$ and $\gamma = 5$.

cdfZ = cdf[5] /. {λ -> 1, γ -> 5};
pdfZ = D[cdfZ, z];
m = NIntegrate[z  pdfZ, {z, 0, ∞}]
(* 0.321508 *)
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