Assortment Planning with Nested Preferences: Dynamic Programming with Distributions as States?

Abstract

The main contribution of this paper is to develop new techniques in approximate dynamic programming, along with the notions of rounded distributions and inventory filtering, to devise a quasi-PTAS for the capacitated assortment planning problem, recently studied by Goyal et al. (Oper Res 64(1):219–235, 2016). Motivated by real-life applications, their nested preference lists model stands as the only setting in dynamic assortment optimization where provably $\epsilon$-optimal inventory levels can be efficiently computed. However, these findings crucially depend on certain distributional assumptions, leaving the general problem wide open in terms of approximability prior to this work. In addition to proposing the first rigorous approach for handling the nested preference lists model in its utmost generality, from a technical perspective, we augment the existing literature on dynamic programming with a number of promising ideas. These are novel algorithmic tools for efficiently keeping approximate distributions as part of the state description, while losing very little information and while accumulating only small approximation errors throughout the overall computation. From a conceptual perspective, at the cost of losing an $\epsilon$-factor in optimality, we show how to dramatically improve on the truly exponential nature of standard dynamic programs, which seem essential for the purpose of computing optimal inventory levels.

Additional Proofs and Details

1.1 Input Specification and Length

For the purpose of studying general distributions from a computational standpoint, we have to make an elementary assumption regarding how the random variable M is specified. To this end, we assume that the number of customers is drawn from a distribution with finite support, say over the set of integers $\{ 0, \ldots , m \}$. In addition, we assume that the latter distribution is given explicitly, i.e., as part of the input specification, the probability for having $\ell$ customers arrive, $\mathrm{Pr} [ M = \ell ]$, is listed explicitly for every $0 \le \ell \le m$. For this reason, given an instance $\mathcal{I}$, reasonable representations are of size

$$\begin{aligned} | \mathcal{I} | = \mathrm {poly}\left( n,m,\log p_n, \log \frac{ 1 }{ \alpha _n }, \log \frac{ 1 }{ \varphi } \right) \ . \end{aligned}$$

Here, $p_n$ is the per-unit selling price of the most expensive product, $\alpha _n = \mathrm{Pr} [ L= (1, \ldots , n) ]$ stands for the consumption probability of this product, and $\varphi$ is the smallest non-zero probability of any value taken by M. Note that the capacity bound C vanishes from this expression, since we can assume without loss of generality that $C \le m$. This follows by observing that there is no point in stocking more than m units, since other than the first m units (counting by increasing product index), all subsequent units will never be consumed.

It is worth mentioning that, even though Goyal et al. [11] appear at first glance to have avoided an explicit representation of M, it is actually kept hidden within an alternative assumption, stating that we have at our possession an “evaluation oracle”. The latter is a procedure that, given any feasible inventory vector $(u_1, \ldots , u_n)$, returns its expected revenue, ${\mathbb E} [ R_M(u_1, \ldots ,u_n) ]$. Their implementation of this procedure makes an explicit use of the probabilities $\mathrm{Pr} [ M = \ell ]$, over all $0 \le \ell \le m$, and runs in $O(Cm^2)$ time.

1.2 Proof of Lemma 1

We begin by identifying an upper bound on the price of product $i_2$. To this end, let $U_2$ be an inventory vector that stocks a single unit of product $i_2$ and nothing more. Then,

$$\begin{aligned} {\mathbb E} \left[ R_M(U^*) \right]\ge & {} {\mathbb E} \left[ R_M(U_2) \right] \\= & {} \mathrm{Pr} \left[ M> 0 \right] \cdot {\mathbb E} \left[ R_M(U_2) | M> 0 \right] \\\ge & {} \mathrm{Pr} \left[ M > 0 \right] \cdot p_{i_2} \alpha _{i_2} \ . \end{aligned}$$

The first inequality holds due to the optimality of $U^*$. The second inequality holds since the probability to consume the single unit of product $i_2$, given that $M > 0$, is greater or equal to the analogous probability when $M=1$, which is exactly $\alpha _{i_2}$. Therefore,

$$\begin{aligned} p_{i_2} \le \frac{ {\mathbb E} \left[ R_M(U^*) \right] }{ \mathrm{Pr} \left[ M > 0 \right] \cdot \alpha _{i_2} } \ . \end{aligned}$$

(8)

Similarly, letting $U_1$ be an inventory vector that stocks a single unit of product $i_1$ and nothing more, we can derive a lower bound on the price of product $i_1$ as follows:

$$\begin{aligned} \frac{ \epsilon }{ \Vert U^* \Vert } \cdot {\mathbb E} \left[ R_M(U^*) \right]\le & {} {\mathbb E} \left[ R_M(U_1) \right] \\= & {} \mathrm{Pr} \left[ M> 0 \right] \cdot {\mathbb E} \left[ R_M(U_1) | M> 0 \right] \\\le & {} \mathrm{Pr} \left[ M> 0 \right] \cdot {\mathbb E} \left[ R_M(U_1) | M = m \right] \\= & {} \mathrm{Pr} \left[ M > 0 \right] \cdot p_{i_1} (1 - (1 - \alpha _{i_1})^m) \ . \end{aligned}$$

The first inequality holds since the first unit of product $i_1$ survived the $\epsilon$-filtering procedure, so its expected revenue in $U^*$ is at least $\frac{ \epsilon }{ \Vert U^* \Vert } \cdot {\mathbb E} [ R_M(U^*) ]$, and the probability to consume this unit in $U_1$ is at least as large as that in $U^*$. Consequently,

$$\begin{aligned} p_{i_1} \ge \frac{ \epsilon \cdot {\mathbb E} \left[ R_M(U^*) \right] }{ \Vert U^* \Vert \cdot \mathrm{Pr} \left[ M > 0 \right] \cdot (1 - (1 - \alpha _{i_1})^m) } \ . \end{aligned}$$

(9)

We conclude by observing that the required upper bound on $\frac{ p_{i_2} }{ p_{i_1} }$ is obtained by combining Eqs. (8) and (9).

1.3 Proof of Lemma 2

We begin by establishing a lower bound on ${\mathbb E} [ R_M( \tilde{U} | M=1 ) ]$. To this end, consider the first unit stocked in $\tilde{U}$. Since this unit survived the $\epsilon$-filtering procedure, we have

$$\begin{aligned} \frac{ \epsilon }{ \Vert U^* \Vert } \cdot {\mathbb E} \left[ R_M(U^*) \right]\le & {} {\mathbb E} \left[ R_M^1( \tilde{U} ) \right] \\= & {} \sum _{\ell = 1}^m \mathrm{Pr} \left[ M = \ell \right] \cdot {\mathbb E} \left[ \left. R_M^1( \tilde{U} ) \right| M = \ell \right] \\\le & {} \sum _{\ell = 1}^m \mathrm{Pr} \left[ M = \ell \right] \cdot \ell \cdot {\mathbb E} \left[ \left. R_M^1( \tilde{U} ) \right| M = 1 \right] \\= & {} {\mathbb E} \left[ \left. R_M^1( \tilde{U} ) \right| M = 1 \right] \cdot {\mathbb E} \left[ M \right] \ . \end{aligned}$$

The second inequality holds since, if the first unit in $\tilde{U}$ appears on product i, then ${\mathbb E} [ R_M^1( \tilde{U} ) | M = 1 ] = p_i \alpha _i$ whereas ${\mathbb E} [ R_M^1( \tilde{U} ) | M = \ell ] = p_i \cdot \sum _{k = 1}^\ell (1 - \alpha _i)^{k-1} \alpha _i \le \ell p_i \alpha _i$. Therefore,

$$\begin{aligned} {\mathbb E} \left[ \left. R_M( \tilde{U} ) \right| M=1 \right] = {\mathbb E} \left[ \left. R_M^1( \tilde{U} ) \right| M = 1 \right] \ge \frac{ \epsilon }{ \Vert U^* \Vert \cdot {\mathbb E} \left[ M \right] } \cdot {\mathbb E} \left[ R_M(U^*) \right] \ . \end{aligned}$$

(10)

Now, for every product i that is stocked in $\bar{U}$ (that is, $\bar{u}_i > 0$), let $U_i$ be an inventory vector that stocks a single unit of product i and nothing more. Then,

$$\begin{aligned} {\mathbb E} \left[ R_M(U^*) \right]\ge & {} {\mathbb E} \left[ R_M( U_i ) \right] \\= & {} \sum _{\ell = 1}^m \mathrm{Pr} \left[ M = \ell \right] \cdot {\mathbb E} \left[ R_M( U_i ) | M = \ell \right] \\\ge & {} \sum _{\ell = 1}^m \mathrm{Pr} \left[ M = \ell \right] \cdot \frac{ \ell }{ m } \cdot {\mathbb E} \left[ R_M( U_i ) | M = m \right] \\= & {} \frac{ 1 }{ m } \cdot {\mathbb E} \left[ R_M( U_i ) | M = m \right] \cdot {\mathbb E} \left[ M \right] \ . \end{aligned}$$

The first inequality holds due to the optimality of $U^*$. To understand the second inequality, note that the probability of consuming a single unit of product i, given $\ell$ customers, is precisely $1 - (1 - \alpha _i)^\ell$, so

$$\begin{aligned} \frac{ {\mathbb E} [ R_M( U_i ) | M = \ell ] }{ {\mathbb E} [ R_M( U_i ) | M = m ] } = \frac{ p_i ( 1 - (1 - \alpha _i)^\ell ) }{ p_i ( 1 - (1 - \alpha _i)^m ) } = \frac{ 1 - x }{ 1 - x^{ m / \ell } } \ge \frac{ \ell }{ m } \ , \end{aligned}$$

where $x = (1 - \alpha _i)^\ell$. The important observation is that the function $1 - x^{ m / \ell }$ is concave (note that $\frac{ m }{ \ell } \ge 1$), so it lies below all of its tangents. One of these tangents passes at $x = 1$, where its equation is exactly $\frac{ m }{ \ell }(1 - x)$. Based on this sequence of inequalities, we have

$$\begin{aligned} {\mathbb E} \left[ R_M( U_i ) | M = m \right] \le \frac{ m }{ {\mathbb E} \left[ M \right] } \cdot {\mathbb E} \left[ R_M(U^*) \right] \ . \end{aligned}$$

(11)

We conclude the proof by splitting ${\mathbb E} [ R_M( \bar{U} ) | M=m ]$ into expected revenues from different products. To this end, let $V_i$ be an inventory vector that stocks $\bar{u}_i$ units of product i and nothing more. Then,

$$\begin{aligned} {\mathbb E} \left[ R_M( \bar{U} ) | M=m \right]\le & {} \sum _{i : \bar{u}_i> 0} {\mathbb E} \left[ R_M( V_i ) | M=m \right] \\\le & {} \sum _{i : \bar{u}_i> 0} \Vert V_i \Vert \cdot {\mathbb E} \left[ R_M( U_i ) | M=m \right] \\\le & {} \frac{ m }{ {\mathbb E} \left[ M \right] } \cdot {\mathbb E} \left[ R_M(U^*) \right] \cdot \sum _{i : \bar{u}_i > 0} \bar{u}_i \\= & {} \frac{ m \cdot \Vert \bar{U} \Vert }{ {\mathbb E} \left[ M \right] } \cdot {\mathbb E} \left[ R_M(U^*) \right] \\\le & {} \frac{ m \cdot \Vert \bar{U} \Vert \cdot \Vert U^* \Vert }{ \epsilon } \cdot {\mathbb E} \left[ \left. R_M( \tilde{U} ) \right| M=1 \right] \ . \end{aligned}$$

The first inequality holds since the expected revenue from product i in $\bar{U}$ can be upper bounded by the expected revenue of this product in $V_i$, as in the latter vector we do not stock any other product. The second inequality follows from observing that, in $V_i$, the probability to consume the first unit (of product i) is greater or equal to the probability to consume the second unit, which is turn is greater or equal to that of the third unit, so forth and so on. In the third and forth inequalities, we simply apply Eqs. (11) and (10), respectively.

1.4 Proof of Lemma 3

To establish the claim, it is sufficient to show that

$$\begin{aligned} \mathrm{Pr} \left[ \left[ X_{\mathcal{A}} - Y \right] ^+ \ge k \right] \ge \mathrm{Pr} \left[ \left[ X-Y \right] ^+_{\mathcal{A}} \ge k \right] \ , \end{aligned}$$

for every $k \in \mathcal{A}$, since the random variable $[ X-Y ]^+_{\mathcal{A}}$ accumulates probability only over values in $\mathcal{A}$. To this end, for an integer $\ell$, let $\max (\ell , \mathcal{A})$ and $\min (\ell , \mathcal{A})$ be the maximum and minimum integers within the subinterval (in the partition $\mathcal{P}_\mathcal{A} [0,m]$) that contains $\ell$, respectively. With these definitions, note that

$$\begin{aligned} \mathrm{Pr} \left[ \left[ X_{\mathcal{A}} - Y \right] ^+ \ge k \right]= & {} \mathrm{Pr} \left[ X_{\mathcal{A}} \ge Y+k \right] \\= & {} \mathrm{Pr} \left[ X_{\mathcal{A}} \ge \max ( Y+k, \mathcal{A} ) \right] \\= & {} \mathrm{Pr} \left[ X \ge \min ( Y+k, \mathcal{A} ) \right] \ . \end{aligned}$$

On the other hand, we have

$$\begin{aligned} \mathrm{Pr} \left[ \left[ X-Y \right] ^+_{\mathcal{A}} \ge k \right] = \mathrm{Pr} \left[ \left[ X-Y \right] ^+ \ge k \right] = \mathrm{Pr} \left[ X \ge Y+k \right] . \end{aligned}$$

In the first equality we used the fact that $k = \max (k, \mathcal{A})$, since $k \in \mathcal{A}$. To conclude, the claim follows by observing that $\min ( Y+k, \mathcal{A} ) \le _{\mathrm {st}}Y+k$, due to the independence of X and Y and the closure under convolutions of the usual stochastic order (see, for instance, [31, Thm. 1.A.3]).

1.5 Proof of Lemma 4

To obtain an upper bound on the above ratio, we compute its numerator by conditioning on the number of units consumed by the first $\lfloor \epsilon \ell \rfloor$ customers. To this end, for $0 \le r \le \Vert \tilde{U} \Vert$, let $\mathcal{A}_r$ stand for the event where exactly r units are consumed by the first $\lfloor \epsilon \ell \rfloor$ customers. Using this notation, we have

$$\begin{aligned}&{\mathbb E} \left[ \left. R_M( \tilde{U} ) \right| M = \lfloor (1+\epsilon )\ell \rfloor \right] \\&\quad = \sum _{r = 0}^{ \Vert \tilde{U} \Vert } \mathrm{Pr} \left[ \mathcal{A}_r | M= \lfloor (1+\epsilon )\ell \rfloor \right] \cdot {\mathbb E} \left[ \left. R_M( \tilde{U} ) \right| M= \lfloor (1+\epsilon )\ell \rfloor \wedge \mathcal{A}_r \right] \\&\quad = \mathrm{Pr} \left[ \mathcal{A}_0 | M= \lfloor (1+\epsilon )\ell \rfloor \right] \cdot {\mathbb E} \left[ \left. R_M( \tilde{U} ) \right| M= \ell \right] \\&\quad \quad + \mathrm{Pr} \left[ \mathcal{A}_1 | M= \lfloor (1+\epsilon )\ell \rfloor \right] \cdot \left( p_{\pi (1)} + {\mathbb E} \left[ \left. R_M( \tilde{U}_{-1} ) \right| M= \ell \right] \right) \\&\quad \quad + \sum _{r = 2}^{ \Vert \tilde{U} \Vert } \mathrm{Pr} \left[ \mathcal{A}_r | M= \lfloor (1+\epsilon )\ell \rfloor \right] \cdot \left( \sum _{k = 1}^r p_{\pi (k)} + {\mathbb E} \left[ \left. R_M( \tilde{U}_{-r} ) \right| M= \ell \right] \right) \ . \end{aligned}$$

In the above equation, $\pi (k)$ denotes the product to which the k-th global unit in $\tilde{U}$ belongs (with price $p_{ \pi (k) }$), while $\tilde{U}_{-r}$ is the inventory vector that results from deleting the first r units in $\tilde{U}$. We proceed by considering these three terms, corresponding to $\mathcal{A}_0$, $\mathcal{A}_1$, and the remaining events $\mathcal{A}_2, \ldots , \mathcal{A}_{ \Vert \tilde{U} \Vert }$. Clearly, the result of dividing the first term by ${\mathbb E} [ R_M( \tilde{U} ) | M= \ell ]$ is upper bounded by 1. For the other two terms, we provide separate bounds as follows.

For the second term,

$$\begin{aligned}&\frac{ \mathrm{Pr} \left[ \mathcal{A}_1 | M= \lfloor (1+\epsilon )\ell \rfloor \right] \cdot ( p_{\pi (1)} + {\mathbb E} [ R_M( \tilde{U}_{-1} ) | M= \ell ] ) }{ {\mathbb E} [ R_M( \tilde{U} ) | M= \ell ] } \\&\qquad \qquad \le \frac{ \lfloor \epsilon \ell \rfloor \alpha _{ \pi (1) } \cdot ( p_{\pi (1)} + {\mathbb E} [ R_M( \tilde{U}_{-1} ) | M= \ell ] ) }{ {\mathbb E} [ R_M( \tilde{U} ) | M= \ell ] } \\&\qquad \qquad \le \frac{ \epsilon \ell \alpha _{ \pi (1) } p_{\pi (1)} }{ {\mathbb E} [ R_M^1( \tilde{U} ) | M= \ell ] } + \ell \alpha _{ \pi (1) } \cdot m \cdot \Vert \tilde{U}_{-1} \Vert \cdot \Vert U^* \Vert \\&\qquad \qquad \le \frac{ \epsilon \ell \alpha _{ \pi (1) } p_{\pi (1)} }{ p_{\pi (1)} (1 - (1 - \alpha _{ \pi (1) })^\ell ) } + m^2 C^2 \varDelta \\&\qquad \qquad \le 2 \epsilon + m^2 C^2 \varDelta \ . \end{aligned}$$

The second inequality holds since $[R_M( \tilde{U} ) | M= \ell ] \ge _{\mathrm {st}}[R_M^1( \tilde{U} ) | M= \ell ]$ for the first summand, and by applying Lemma 2 for the second summand, noting that $\tilde{U}_{-1} \le \tilde{U}$. The third inequality holds since $\ell \le m$, $\Vert \tilde{U}_{-1} \Vert \le \Vert U^* \Vert \le C$, and $\alpha _{ \pi (i) } \le \varDelta$. The fourth inequality follows from observing that

$$\begin{aligned} 1 - \left( 1 - \alpha _{ \pi (1) } \right) ^\ell \ge 1 - e^{ -\alpha _{ \pi (1) } \ell } \ge \frac{ \alpha _{ \pi (1) } \ell }{ 2 } \ , \end{aligned}$$

(12)

where the last inequality holds since $e^{-x} \le 1 - \frac{ x }{ 2 }$ over the interval [0, 1]. Note that we indeed have $\alpha _{ \pi (1) } \ell \le 1$, since by assumption, $\alpha _{ \pi (1) } \le \varDelta \le \frac{ 1 }{ m }$.

Now, regarding the third term, for any $2 \le r \le \Vert \tilde{U} \Vert$,

$$\begin{aligned}&\frac{ \mathrm{Pr} \left[ \mathcal{A}_r | M= \lfloor (1+\epsilon )\ell \rfloor \right] \cdot ( \sum _{k = 1}^r p_{\pi (k)} + {\mathbb E} [ R_M( \tilde{U}_{-r} ) | M= \ell ] ) }{ {\mathbb E} [ R_M( \tilde{U} ) | M= \ell ] } \\&\qquad \qquad \le \frac{ \lfloor \epsilon \ell \rfloor ^r \cdot ( \prod _{k = 1}^r \alpha _{ \pi (k) } ) \cdot ( r p_{\pi (r)} + {\mathbb E} [ R_M( \tilde{U}_{-r} ) | M= \ell ] ) }{ {\mathbb E} [ R_M( \tilde{U} ) | M= \ell ] } \\&\qquad \qquad \le \frac{ r \cdot \prod _{k = 1}^r \epsilon \ell \alpha _{ \pi (k) } }{ {\mathbb E} [ R_M( \tilde{U} ) | M= \ell ] } \cdot \frac{ \Vert U^* \Vert }{ \epsilon \alpha _{ \pi (r)} } \cdot \left( 1 - \left( 1 - \alpha _{\pi (1)} \right) ^m \right) \cdot p_{\pi (1)} \\&\qquad \qquad \quad + \left( \prod _{k = 1}^r \epsilon \ell \alpha _{ \pi (k) } \right) \cdot \frac{ m \cdot \Vert \tilde{U}_{-r} \Vert \cdot \Vert U^* \Vert }{ \epsilon } \\&\qquad \qquad \le r \ell \cdot \Vert U^* \Vert \cdot \left( \prod _{k = 1}^{r-1} \epsilon \ell \alpha _{ \pi (k) } \right) \cdot \frac{ ( 1 - (1 - \alpha _{\pi (1)} )^m) \cdot p_{\pi (1)} }{ {\mathbb E} [ R_M^1( \tilde{U} ) | M= \ell ] } \\&\qquad \qquad \quad + \frac{ m C^2 }{ \epsilon } \cdot \prod _{k = 1}^r \epsilon \ell \alpha _{ \pi (k) } \\&\qquad \qquad = r \ell \cdot \Vert U^* \Vert \cdot \left( \prod _{k = 1}^{r-1} \epsilon \ell \alpha _{ \pi (k) } \right) \cdot \frac{ ( 1 - (1 - \alpha _{\pi (1)} )^m) \cdot p_{\pi (1)} }{ (1 - (1 - \alpha _{\pi (1)} )^\ell ) \cdot p_{\pi (1)} }\\&\qquad \qquad \quad + \frac{ m C^2 }{ \epsilon } \cdot \prod _{k = 1}^r \epsilon \ell \alpha _{ \pi (k) } \\&\qquad \qquad \le r \ell \cdot \Vert U^* \Vert \cdot \left( \prod _{k = 1}^{r-1} \epsilon \ell \alpha _{ \pi (k) } \right) \cdot \frac{ m \alpha _{ \pi (1) } }{ \ell \alpha _{ \pi (1) } / 2 } + \frac{ m C^2 }{ \epsilon } \cdot \prod _{k = 1}^r \epsilon \ell \alpha _{ \pi (k) } \\&\qquad \qquad \le 2 m C r \prod _{k = 1}^{r-1} \epsilon \ell \alpha _{ \pi (k) } + \frac{ m C^2 }{ \epsilon } \cdot \prod _{k = 1}^r \epsilon \ell \alpha _{ \pi (k) } \\&\qquad \qquad \le 3 m^2 C^2 (\epsilon \ell \varDelta )^{ r-1 } \ . \end{aligned}$$

The second inequality follows from applying Lemma 1 to bound $p_{\pi (r)}$ in the first summand, and from applying Lemma 2 in the second summand. The third inequality holds since $[R_M( \tilde{U} ) | M= \ell ] \ge _{\mathrm {st}}[R_M^1( \tilde{U} ) | M= \ell ]$ for the first summand, and since $\Vert \tilde{U}_{-r} \Vert \le \Vert U^* \Vert \le C$ for the second summand. The fourth inequality holds since $1 - (1 - \alpha _{\pi (1)} )^m \le m \alpha _{\pi (1)}$, due to the convexity of the LHS, and since $1 - (1 - \alpha _{\pi (1)} )^\ell \ge \frac{ \ell \alpha _{\pi (1)} }{ 2 }$ by Eq. (12). It follows that the entire third term can be upper bounded by

$$\begin{aligned} 3 m^2 C^2 \cdot \sum _{r = 2}^{ \Vert \tilde{U} \Vert } (\epsilon \ell \varDelta )^{ r-1 } \le 3 m^3 C^3 \varDelta \ . \end{aligned}$$

1.6 Lower Bound on Non-zero Probabilities

In what follows, suppose that $(u_1, \ldots , u_n)$ is a feasible inventory vector, and let $M_i$ be the incoming traffic to product i, that is, the number of customers remaining once product $i-1$ is completed consumed. As previously mentioned, $M_i = [M - \sum _{j = 1}^{i-1} Y_j]^+$, where $Y_j \sim \mathrm {NB}(u_j, \alpha _j)$. The next lemma provides a lower bound for the following question: How small can $\mathrm{Pr} [ M_i = \ell ]$ get, assuming that this probability is strictly positive?

Lemma 5

For every $1 \le i \le n$ and $1 \le \ell \le m$, we have either $\mathrm{Pr} [ M_i = \ell ] = 0$ or

$$\begin{aligned} \mathrm{Pr} \left[ M_i = \ell \right] \ge \varphi \cdot \prod _{j = 1}^{i-1} \left( \alpha _j^{ u_j } (1-\alpha _j)^{ m-u_j } \right) \ , \end{aligned}$$

where $\varphi$ is the smallest non-zero probability of any value taken by M, i.e.,

$$\begin{aligned} \varphi = \min _{ 1 \le j \le m} \left\{ \mathrm{Pr} [ M = j ] : \mathrm{Pr} [ M = j ] > 0 \right\} \ . \end{aligned}$$

Proof

The proof proceeds by induction on the product index i, noting that the base case $i = 1$ follows directly from the definition of $\varphi$. So, consider some $i \ge 2$ and $1 \le \ell \le m$, for which $\mathrm{Pr} [ M_i = \ell ] > 0$. By conditioning on $M_{i-1}$, we have

$$\begin{aligned} \mathrm{Pr} \left[ M_i = \ell \right]= & {} \sum _{k = \ell }^m \mathrm{Pr} \left[ M_{i-1} = k \right] \cdot \mathrm{Pr} \left[ M_i = \ell | M_{i-1} = k \right] \\= & {} \sum _{k = \ell }^m \mathrm{Pr} \left[ M_{i-1} = k \right] \cdot \mathrm{Pr} \left[ Y_{i-1} = k - \ell \right] \ . \end{aligned}$$

Therefore, there exists some integer $\bar{k}$, for which the inner term is positive. We now use the fact that $Y_{i-1} \sim \mathrm {NB}(u_{i-1}, \alpha _{i-1})$ along with the induction hypothesis, to obtain

$$\begin{aligned}&\mathrm{Pr} \left[ M_i = \ell \right] \ge \mathrm{Pr} \left[ M_{i-1} = \bar{k} \right] \cdot \mathrm{Pr} \left[ Y_{i-1} = \bar{k} - \ell \right] \\&\qquad \ge \varphi \cdot \prod _{j = 1}^{i-2} \left( \alpha _j^{ u_j } (1-\alpha _j)^{ m-u_j } \right) \cdot {{\bar{k} - \ell - 1} \atopwithdelims (){{u}_{i-1} - 1}} \alpha _{i-1}^{ {u}_{i-1} } \left( 1 - \alpha _{i-1}\right) ^{ \bar{k} - \ell - {u}_{i-1} } \\&\qquad \ge \varphi \cdot \prod _{j = 1}^{i-2} \left( \alpha _j^{ u_j } (1-\alpha _j)^{ m-u_j } \right) \cdot \left( \alpha _{i-1}^{ {u}_{i-1} } \left( 1 - \alpha _{i-1}\right) ^{ m - {u}_{i-1} } \right) \\&\qquad = \varphi \cdot \prod _{j = 1}^{i-1} \left( \alpha _j^{ u_j } (1-\alpha _j)^{ m-u_j } \right) \ . \end{aligned}$$

$\square$