2. Bounds on P(N > n)

Proposition 2.1.

\begin{equation*}P(N \gt n) \geq P(S_1 \lt s_1) \prod_{k=2}^n P(S_k \lt s_k | S_{k-1} \lt s_{k-1}).\end{equation*}

To establish Proposition 2.1, we present some lemmas. The first one is Efron’s theorem[Reference Efron3].

Lemma 2.2. Efron’s Theorem

If $X_1, \ldots, X_r$ are independent log-concave random variables, then $(X_1, \ldots, X_r) \,|\, \sum_{t=1}^r X_t = x$ is stochastically increasing in $x.$ That is, for any component-wise increasing function $g(x_1, \ldots, x_r)$

\begin{equation*}E\big [g(X_1, \ldots, X_r) | \sum_{t=1}^r X_t = x\big ] \;\;\mbox{is increasing in}\;\;x\end{equation*}

Our next lemma states that S_k conditional on $S_1 \lt s_1, \ldots, S_{k} \lt s_k $ is likelihood ratio smaller than S_k conditional on $S_k \lt s_k.$

Lemma 2.3.

\begin{equation*}S_k| (S_1 \lt s_1, \ldots, S_{k} \lt s_k) \; \leq_{lr} \; S_k| S_{k} \lt s_k.\end{equation*}

Proof. We need to show that the ratio of the conditional density of S_k given $S_1 \lt s_1, \ldots, S_{k} \lt s_k$ to the conditional density of S_k given $ S_{k} \lt s_k$ is decreasing. Now, for $t \leq s_k$

\begin{eqnarray*} f_{S_k| S_1 \lt s_1, \ldots, S_{k} \lt s_k}(t)\!\! &=& \!\!\frac{ f_{S_k}(t) P(S_1 \lt s_1, \ldots, S_{k} \lt s_k|S_k = t)}{P( S_1 \lt s_1, \ldots, S_{k} \lt s_k)} \\ f_{S_k| S_{k} \lt s_k}(t)\!\! &=& \!\!\frac{ f_{S_k}(t) P( S_{k} \lt s_k|S_k = t)}{P( S_{k} \lt s_k)} = \frac{ f_{S_k}(t) }{P( S_{k} \lt s_k)} \end{eqnarray*}

Hence we need to show that $ P(S_1 \lt s_1, \ldots, S_{k} \lt s_k|S_k = t) $ is a decreasing function of t. However, this follows from Efron’s theorem because $ g(x_1, \ldots, x_k) = 1- I\{x_1 \lt s_1, x_1 + x_2 \lt s_2, \ldots, x_1+ \ldots + x_k \lt s_k\} $ is an increasing function of $(x_1, \ldots, x_k)$.

Lemma 2.4.

\begin{equation*}P(S_k \lt s_k | S_1 \lt s_1, \ldots, S_{k-1} \lt s_{k-1} ) \geq P(S_k \lt s_k | S_{k-1} \lt s_{k-1})\end{equation*}

Proof. Proposition 2.1.

Proposition 2.1 follows from Lemma 2.4 upon using that

\begin{eqnarray*} P(N \gt n) &=& P(S_1 \lt s_1, \ldots, S_n \lt s_n ) \\ &=& P(S_1 \lt s_1) \prod_{k=2}^n P(S_k \lt s_k | S_1 \lt s_1, \ldots, S_{k-1} \lt s_{k-1}) \end{eqnarray*}

Proposition 2.1 yields the following lower bound on $E[N]$.

Corollary 2.5.

\begin{eqnarray*} E[N] &=& \sum_{n=0}^{\infty} P(N \gt n) \\ & \geq & 1 + P(S_1 \lt s_1) \big (1 + \sum_{n=2}^{\infty} \prod_{k=2}^n P(S_k \lt s_k|S_{k-1} \lt s_{k-1})\big ) \end{eqnarray*}

Remark. The log concave condition is essential for establishing Proposition 2.1. For a counterexample, suppose that $p_0 = \epsilon,\; p_2 = .2, \; p_5 = .8 - \epsilon,$ where $p_j = P(X_i = j)$ and ϵ is a small positive number. Then

\begin{equation*} P(S_3 \lt 6.5|S_1 \lt 1, S_2 \lt 5.5) = P(X_2 + X_3 \lt 6.5) \approx .04\end{equation*}

whereas

\begin{equation*} P(S_3 \lt 6.5| S_2 \lt 5.5) \approx .2\end{equation*}

The conditional expectation inequality (see [Reference Ross6]) can be used to obtain an upper bound on $P(N \gt n).$

Lemma 2.7. The Conditional Expectation Inequality

For events $B_1, \ldots, B_n$

\begin{equation*}P( \cup_{i=1}^n B_i) \geq \sum_{i=1}^n \frac{P(B_i)}{1 + \sum_{j \neq i} P(B_j|B_i)} .\end{equation*}

With $B_i = A_i^c = \{S_i \geq s_i\}, $ the inequality yields that

\begin{equation*}P(N \leq n) \geq \sum_{i=1}^n \frac{P^2(B_i) } {P(B_i) + \sum_{j \neq i} P(B_i B_j) } \end{equation*}

Whereas for many logconcave distributions it is difficult to compute $P(S_k \lt s_k | S_{k-1} \lt s_{_{k-1}})$, this is easily accomplished in important special cases such as normal, exponential, binomial, and Poisson. Example 2.8 considers the normal case and Example 2.9 the exponential case.

Example 2.8. Suppose the X_i are normal random variables with mean µ and variance 1. Let Z be a standard normal whose distribution function is Φ; let U be uniform on $(0, 1);$ and let $c_{n-1} = \frac{s_{n-1} - (n-1)\mu}{\sqrt{n-1}}.$ Because $S_{n-1}$ is normal with mean $(n-1) \mu$ and variance $n-1,$ it follows that

\begin{eqnarray*} S_{n-1} | S_{n-1} \lt s_{n-1}\!\! & =_{st} & \!\!(n-1) \mu + \sqrt{n-1} \,Z | \;Z \lt c_{n-1} \\ \!\!& =_{st} & \!\!(n-1) \mu + \sqrt{n-1} \, \Phi^{-1}(U) \; | \; \Phi^{-1}(U) \lt c_{n-1} \\ \!\!& =_{st} & \!\!( n-1) \mu + \sqrt{n-1} \, \Phi^{-1}(U) \; | \; U \lt \Phi(c_{n-1} ) \\ \!\!& =_{st} & \!\!( n-1) \mu + \sqrt{n-1}\, \Phi^{-1}(U\Phi(c_{n-1} ) ) \end{eqnarray*}

Hence

\begin{eqnarray*} P(S_n \lt s_n | S_{n-1} \lt s_{n-1}) &=& \!E[ E[ I\{S_n \lt s_n \} | S_{n-1} ] | S_{n-1} \lt s_{n-1} ] \\ \!&=& \!E[ \Phi( s_n - \mu - S_{n-1} ) | S_{n-1} \lt s_{n-1} ] \\ \!&=& \!E[ \Phi( s_n - n \mu - \sqrt{n-1} \Phi^{-1}(U \Phi(c_{n-1})) ] \\ \!&=& \!\int_0^1 g(x) dx, \end{eqnarray*}

where $\;g(x) = \Phi \left( s_n - n \mu - \sqrt{n-1} \, \Phi^{-1}(x \Phi(c_{n-1}) \right).$ Because Φ and Φ⁻¹ are both increasing functions, it follows that g(x) is a decreasing function of x. Since $\;\int_0^1 g(x) dx = \sum_{i=1}^r \int_{(i-1)/r}^{i/r} g(x) dx ,$ this shows that, for any $r,$

\begin{equation*}\frac{1}{r} \sum_{i=1}^r g( \frac{i-1}{r}) \geq \int_0^1 g(x) dx \geq \frac{1}{r} \sum_{i=1}^r g(\frac{i}{r})\end{equation*}

To utilize the conditional expectation inequality we need to compute $P(S_i \gt s_i, S_j \gt s_j), i \neq j.$ To do so, suppose that i < j, and let $c_i = \frac{s_{i} - i\mu}{\sqrt{i}}$. Then, with $\bar{\Phi} = 1 - \Phi,$ arguing as before yields

\begin{eqnarray*} S_{i} | S_{i} \gt s_{i} & =_{st} & i \mu + \sqrt{i} \; \Phi^{-1}(U) \; | \; U \gt \Phi(c_i ) \\ & =_{st} & i \mu + \sqrt{i}\; \Phi^{-1} \left(\Phi(c_i ) + \bar{\Phi}(c_i) U \right) \end{eqnarray*}

Using that $S_j |S_i$ is normal with mean $S_i + (j-i) \mu$ and variance $(j-i)^2$ yields that

\begin{eqnarray*} P(S_j \gt s_j|S_i \gt s_i)\! &=& \!E\big[ E[ I\{S_j \gt s_j\} | S_{i} ] | S_{i} \gt s_{i} \big] \\ \!&=& \! E\big[ \bar{\Phi}\big( \frac{s_j - S_i -(j-i)\mu}{\sqrt{j-i}} \big ) | S_{i} \gt s_{i} \big] \\ \!&=& \! E \big [ \bar{\Phi} \left( \frac{s_j - j \mu - \sqrt{i}\; \Phi^{-1} \left(\Phi(c_i ) + \bar{\Phi}(c_i) U \right) }{\sqrt{j-i}} \right) \big ] \\ \!&=& \!\int_0^1 h_{i,j} (x) dx, \end{eqnarray*}

where $h_{i,j}(x) = \bar{\Phi} \left( \frac{s_j - j \mu - \sqrt{i}\; \Phi^{-1} \left(\Phi(c_i ) + \bar{\Phi}(c_i) x \right) }{\sqrt{j-i}} \right) .$ Because $h_{i,j}(x)$ is an increasing function of x, this gives

\begin{equation*} \frac{1}{r} \sum_{i=1}^r h_{i,j} \big(\frac{i}{r}\big ) \geq \int_0^1 h(x) dx \geq \frac{1}{r} \sum_{i=1}^r h_{i,j}\big( \frac{i-1}{r}\big). \end{equation*}

Example 2.9. Suppose the X_i are exponential random variables with rate $\lambda.$ Let N(t) be the number of events by time t of the Poisson process that has X_i as its i ^th interarrival time, $i \geq 1.$ Now, if $s_{n-1} \leq s_n$ then

\begin{eqnarray*} P(S_n \lt s_n, S_{n-1} \lt s_{n-1})\! &=& \!P(N(s_n) \geq n , N(s_{n-1}) \geq n -1) \\ \!&=& \!\sum_{i=n-1}^{\infty} P(N(s_n) \geq n | N(s_{n-1}) = i) P(N(s_{n-1}) = i) \\ \!&=& \!(1 - e^{- \lambda (s_n - s_{n-1})}) P(N(s_{n-1}) = n-1) + \sum_{i=n}^{\infty} P(N(s_{n-1}) = i) \\ \!&=& \!P(N(s_{n-1}) \geq n -1) - e^{- \lambda (s_n - s_{n-1})} P(N(s_{n-1}) = n-1) \end{eqnarray*}

giving that

\begin{equation*} P(S_n \lt s_n| S_{n-1} \lt s_{n-1}) = 1- \frac{e^{- \lambda (s_n - s_{n-1})} P(N(s_{n-1}) = n-1)}{P(N(s_{n-1}) \geq n -1) }, \quad s_{n-1} \leq s_n\end{equation*}

If $s_{n-1} \geq s_n,$ then $\; P(S_n \lt s_n, S_{n-1} \lt s_{n-1}) = P(N(s_n) \geq n ) , $ giving that

\begin{equation*}P(S_n \lt s_n | S_{n-1} \lt s_{n-1}) = \frac{ P(N(s_n) \geq n ) }{P(N(s_{n-1}) \geq n-1)} \end{equation*}

To compute $P(S_i \gt s_i, S_j \gt s_j) = P(N(s_i) \lt i, N(s_j ) \lt j)$ suppose that $i \lt j.$ If $s_j \gt s_i,$ conditioning on $N(s_i)$ yields

\begin{equation*}P(S_i \gt s_i, S_j \gt s_j) = \sum_{r=0}^{i-1} P(N(s_i) = r) P(N(s_j - s_i ) \lt j-r) \end{equation*}

If $s_i \gt s_j,$ then

\begin{equation*}P(S_i \gt s_i, S_j \gt s_j) = P(S_i \gt s_i) = P(N(s_i) \lt i).\end{equation*}

In Tables 1–2 and Figure 1, we present two numerical results for the probability bounds of $P(N \gt n)$. One assumes that X_i follows a normal distribution with mean 1 and variance 1, denoted $X_i \sim N(1,1)$. The other assumes that X_i follows an exponential distribution with rate parameter 1, denoted $X_i \sim Exp(1)$. The boundary $s_k = 2 + 2 \sqrt{k},\, k \geq 1$. For values of n ranging from 2 to 20, we calculate the lower and upper bounds for $P(N \gt n)$ using the analytical method described alongside Monte Carlo simulation estimates.

Figure 1. Probability bounds and Monte Carlo estimates for $P(N \gt n). s_k = 2 + 2\sqrt{k}$, Left: $X_i \sim N(1,1)$. Right: $X_i \sim \mathrm{Exp}(1)$.

Table 1. Probability bounds and Monte Carlo estimates for $P(N \gt n)$ with $X_i \sim N(1,1)$ and $s_k = 2 + 2 \sqrt{k}$ for n = 2 to 20.

Table 2. Probability bounds and Monte Carlo estimates for $P(N \gt n)$ with $X_i \sim Exp(1)$ and $s_k = 2 + 2 \sqrt{k}$ for n = 2 to 20

Whereas the bounds on $P(N \gt n)$ also yield bounds on $E[N]$, additional bounds for a square root barrier are given in the next section.

3. Additional bounds on $E[N] $ for a square root barrier

Suppose that $s_k = a + b \sqrt{k}, k \geq 1,$ where $a \geq 0, b \gt 0.$ Also, suppose the log concave random variables X_i have a positive mean. Now, conditional on N and $S_{N-1},$ the random variable S_N is distributed as $a+b \sqrt{N} $ plus the amount by which $X,$ a random variable having density f exceeds the positive value $a+b \sqrt{N} - S_{N-1} $ given that it does exceed that value. But a log concave random variable X conditioned to be positive has an increasing failure rate (see Shaked and Shanthikumar[Reference Moshe and George Shanthikumar.5]) implying that $S_N - (a + b \sqrt{N})$ is stochastically smaller than $X|X \gt 0.$ As this is true no matter what the values of N and $S_{N-1}$, it follows that

\begin{equation*}E[S_N] \leq a + bE[ \sqrt{N} ] + E[X|X \gt 0] \end{equation*}

Using Wald’s equation and Jensen’s inequality the preceding implies that

\begin{equation*}\mu E[N] \leq a + b \sqrt{E[N]} + E[X|X \gt 0] \end{equation*}

With $d = a + E[X|X \gt 0],$ the preceding can be written as

\begin{equation*}\mu E[N] - d \leq b \sqrt{E[N]} \end{equation*}

If $ d \leq \mu E[N] ,$ which can be checked using Corollary 2.5, the preceding yields that

\begin{equation*}\mu^2 E^2[N] + d^2 - (2d \mu+ b^2) E[N] \leq 0\end{equation*}

Because the function $g(x) = \mu^2 x^2 - (2d \mu+ b^2) x + d^2$ is convex with $g(0) \gt 0, \lim_{x \rightarrow \infty} g(x) = \infty$, it follows that $g(x) \lt 0$ in the region between the two roots of $g(x) = 0.$ Thus, $E[N]$ lies between these two roots.

In Table 3, we give the numerical results of the lower and upper bounds of $E[N]$ and compare them with the Monte Carlo estimate of $E[N]$. In this case, $X_i \sim N(1,1)$ and $s_k = a + b \sqrt{k}, k \geq 1$.

Table 3. Comparison of Simulated $E[N]$ with Lower and Upper Bounds for Different Values of a and b.

Let $\hat{E[N]}$ be the Monte Carlo estimate of $E[N]$, and let LB denote the lower bound in Corollary 2.5. Since the smaller root of $ \mu^2 x^2 - (2d \mu+ b^2) x + d^2 = 0$ does not yield a good result, it is excluded. UB₁ is the upper bound by conditional expectation inequality, and UB₂ is the larger root. The results are as follows.