# What values can the process $X_n$ take when $P(Y = k) = a_k = 0$ for $k \geq 2$

A newspaper uses one ton of newsprint every day. It buys its newsprint from a local distributor. This ditributor supplies the newsprint in one-ton rolls at the cheapest price, but unfortunatley its deliveries are erratic. Hence, the number of rolls of newsprint in the newpaper warehouse varies randomly. If on a particular day, there are no rolls in the warehouse, the newspaper buys and uses a rold from an emergency supplier. Let $$Y_n$$ be the number of rolls received from the local distributor on day $$n$$. The supply arrives in the evening, whereas the one tone is consumed in the morning. Let $$X_n$$ be the number of rolls in the warehouse at the beginning of day $$n$$ before the one ton demand has to be satisfied and the supply for day $$n$$ has arrived. Thus $$X_{n=1} = X_n - 1 + Y_n$$ if $$X_n>0$$ and $$X_{n+1} = X_n$$ is $$X_n = 0$$.

Suppose that the random variables $$\{Y_n, n \geq 0\}$$ are idependent of $$X_{0}$$. Besides, suppose that they are naturally independent and identically distributed and $$P(Y_n = k) = a_k, k = 0,1,...$$.

• Suppose $$a_k = 0$$ for $$k \geq 2$$ and $$X_0 \in \{0, 1\}$$, what values can the process $$X_n$$ take in this situation?

From my transition matrix, I said if we are only looking at the values for when $$i, j$$ are $$0$$ and $$1$$, we would get the little $$2 \times 2$$ matrix of the whole one step transition matrix and so this would just be

$$\pmatrix{a_0 & a_1 \\ a_0 & a_1}$$

and so the only values that $$X_n$$ can take in this situation are $$a_0$$ and $$a_1$$. But in the answers it says it should be $$0$$ and $$1$$. Why is this?

If you just look at your recursions then this should yield the solution. If $X_0=0$ then $X_n = 0$ for all $n$ -> let's call this the absorbtion.
So we analyze the case that $X_0=1$. Then $$X_{n+1} = X_n - 1 + Y_n$$ Consider $n=0$ if $Y_0 = 0$ then $X_1 = 1 - 1 + 0 = 0$ and we are caught in $0$ due to the absorption. If $Y_0=1$ then $X_1 = 1 - 1 + 1 = 1$ and we are back in the case $X_1 = 1$. Because $Y$ can only take the values $0$ or $1$ (all other probabilities are zero). Then we can continue this reasoning for the transition from $X_1$ to $X_2$ and finally for $X_n,n\ge 0$ this concludes the proof.
• An when you consider the transition matrix the $a_{i,j}$ are the probabilities - not the values of $X$. The values are ${0,1}$ otherwise you would have a bigger matrix. The dimension is given by the number of states. – Ric Jan 24 '13 at 13:53