I can't understand this difference equation step

Question

I am working on birth-death processes and I can't understand a step that is taken in a proof.

The mean of a process is defined as

$$\mu(t) = \sum_{n=1}^{\infty}np_n(t)$$

At certain stage in the proof we go from

$$\lambda \sum_{n=0}^{\infty}(n+1)p_n(t) + \lambda \sum_{n=1}^{\infty}np_n(t)$$

to

$$=\lambda \mu(t) + \lambda\mu(t)$$

I don't see how this is valid. The expansions for the summations aren't the same. For instance, $1 \times p_0 + 2 \times p_1 + ... \ne 1 \times p_1 + 2 \times p_2 + ...$

Am I missing something here, anyone know what's going on?

Here is the proof

Answer 1

It’s a little confusing to have both a constant $\mu$ and a function $\mathbf{\mu}$, so I’m going to replace the former by $\nu$ for this answer. In addition, there appears to be a typo in the first line: the argument makes sense only if the expression on the righthand side of the first equals sign is

$$\lambda\sum_{n\ge 1}np_{n-1}(t)-\lambda\sum_{n\ge 1}np_n(t)-\nu\sum_{n\ge 1}n^2p_n(t)+\nu\sum_{n\ge 1}n(n+1)p_{n+1}(t)\;,\tag{1}$$

with $n^2p_n(t)$ in the third summation.

Shifting the indices on the first term and using the definition of $\mu(t)$ lets us rewrite the first two terms as

$$\lambda\sum_{n\ge 0}(n+1)p_n(t)-\lambda\mu(t)=\lambda\sum_{n\ge 0}np_n(t)+\lambda\sum_{n\ge 0}p_n(t)-\lambda\mu(t)=\lambda\mu(t)+\lambda-\lambda\mu(t)=\lambda\;.$$

Ignoring the factor of $\nu$, the last term can be rewritten as

$$\begin{align*} \sum_{n\ge 2}(n-1)np_n(t)&=\sum_{n\ge 2}n^2p_n(t)-\sum_{n\ge 2}np_n(t)\\\\ &=\left(\sum_{n\ge 1}n^2p_n(t)-p_1(t)\right)-\left(\sum_{n\ge 1}np_n(t)-p_1(t)\right)\\\\ &=\sum_{n\ge 1}n^2p_n(t)-\sum_{n\ge 1}np_n(t)\\\\ &=\sum_{n\ge 1}n^2p_n(t)-\mu(t)\;. \end{align*}$$

$(1)$ now reduces to

$$\lambda-\nu\sum_{n\ge 1}n^2p_n(t)+\nu\sum_{n\ge 1}n^2p_n(t)-\nu\mu(t)=\lambda-\nu\mu(t)\;,$$

as desired.

Answer 2

Basically, the definition of expected value gives $$\sum_{n=0}^\infty n p_n(t)=\vec{\mu}(t),$$ and the definition of a probability distribution gives $$\sum_{n=0}^\infty p_n(t)=1.$$

In particular, $$\sum_{n=0}^\infty (n+1)p_n(t) = \vec{\mu}(t)+1.$$

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Being explicit: \begin{align*} &\phantom{{}={}} \lambda \sum_{n = 0}^\infty(n+1)p_n(t) - \lambda \vec{\mu}(t) - \mu \sum_{n=1}^\infty n^2 p_n(t) + \mu \sum_{n=2}^\infty n(n-1) p_n(t)\\ &= \lambda (\vec{\mu}(t)+1) - \lambda \vec{\mu}(t) - \mu \sum_{n=1}^\infty n^2 p_n(t) + \mu \sum_{n=1}^\infty n(n-1) p_n(t) & \text{adding a zero addend}\\ &= \lambda - \mu \sum_{n=1}^\infty (n(n-1)-n^2) p_n(t)\\ &= \lambda +\mu \sum_{n=1}^\infty n p_n(t)\\ &= \lambda+ \mu \sum_{n=0}^\infty n p_n(t) & \text{adding a zero addend}\\ &= \lambda+\mu\vec{\mu}(t). \end{align*}