I was reading a paper that gives a dynamic programming model of an R&D project. It said that the performance drift (the uncertainty in the performance of the product being developed) follows a binomial distribution. From period t to the next period, the performance may unexpectedly improve
with probability p, or it may deteriorate with probability 1−p because of unexpected adverse events. We generalize the binomial distribution by allowing the performance improvement and deterioration, respectively, to be “spread” over the next N performance states with transition probabilities.
pij = $p/n$ if $j\in$ {$i+1/2,…,i+n/2$} , $(1-p)/n$ if $j\in$ {$i-1/2,…,i-n/2$}
I understand upto this point.
Now it says that mean of this distribution is $((N + 1)/4)(2p − 1)+i$ and the variance is $((N + 1)/8)(N/3+(N+1)(1/3 −((2p − 1)$2$/2)))$
I have no idea how do they get this mean and variance. From what I know about binomial distribution, mean is np and variance is npq. ($q=1-p$). Can anyone please show how have they derive this mean and variance?
Edit 1: pij means probability of going from product performance i to product performance j. So basically if product performance is iat time t and j at time t+1 and if $j>i$ then product performance improved and vice-versa. See the left section of this figure and the Note underneath to understand the meaning of "spread":
Best Answer
I'll answer how they derived the mean or expected value of the distribution, and then I'll leave it to you to derive the variance (it'll involve a similar sort of method).
This is assuming $i=0$ (you can easily prove that for an arbitrary $i$ it just amounts to adding $i$ at the end). Think of this as just 'shifting' the distribution $i$ units to the right (or left).
For a random variable $X$ we define the 'mean' or expected value of $X$ as
$\sum_{\forall x} P(X=x)x$ where $P(X=x)$ is the probability that the random variable $X$ takes on the value $x$.
Lets first define the random variable in this scenario, here it is:
$X = {\frac{i}{2}}$, $\forall i \in +/- \{1,2,3,...N\}$
We can also define a probability function for $X$ as
$P(X=x) = {\frac{p}{N}}\\ \forall i \in + \{1,2,...,N\}$
$P(X=x) = {\frac{1-p}{N}} \\ \forall i\in - \{1,2,...,N\}$
So
$E(x) = \sum_{\forall x}P(X=x)x = \sum_{i \in \{1,2,...,n\}}\frac{p}{N}\frac{i}{2} + \sum_{i \in \{-1,-2,...,-n\}}\frac{1-p}{N}\frac{i}{2} $
Also,
$\sum_{i \in \{1,2,...,n\}}\frac{p}{N}\frac{i}{2} = \frac{1}{2}\frac{p}{N} \sum_{i \in \{1,2,...,n\}}{i}$
But $ \sum_{i \in \{1,2,...,n\}}{i}$ is just the sum of the first $N$ natural numbers, which is known to be $\LARGE \frac{N(N+1)}{2}$
So $\sum_{i \in \{1,2,...,n\}}\frac{p}{N}\frac{i}{2} = \frac{1}{2}\frac{p}{N} \sum_{i \in \{1,2,...,n\}}{i} = \frac{1}{2}\frac{p}{N}\frac{N(N+1)}{2} = \frac{p(N+1)}{4}$
Similarly, we can also show that $\sum_{i \in \{-1,-2,...,-n\}}\frac{1-p}{N}\frac{i}{2} = \frac{p-1(N+1)}{4}$
And so $E(x) = \sum_{\forall x}P(X=x)x = \sum_{i \in \{1,2,...,n\}}\frac{p}{N}\frac{i}{2} + \sum_{i \in \{-1,-2,...,-n\}}\frac{1-p}{N}\frac{i}{2} = \frac{p(N+1)}{4} + \frac{(p-1)(N+1)}{4} = \frac{2p-1(N+1)}{4}$