# Formula for the variance of a renewal process

Let $N(t)$ be a renewal process, with a sequence of IID inter-arrival times $X_{1}, X_{2}, \dots$ having finite second moment: $EX_{i}^{2} < \infty$.

How would I show that $$\mathrm{Var}N(t)= 2 \int^{t}_{0} m(t-s) \cdot m'(s)ds + m(t) - m(t)^{2}$$

where $m(t) = E[N(t)]$ and can be written $m(t) = F(t) + \int_{0}^{t} m(t-x)f(x)dx$ where $f$ is the density of the inter-arrival times and $F$ is the CDF.

I'd just like to know how this works out since I'm weak in computation. I've tried using the usual identity for variance, but I didn't know if that would lead anywhere.

• Where did you find this formula? In the case that $X_1 = 1$ almost surely, we have $\text{Var}(N_t) = 0$ and $m(t) = \lfloor t \rfloor$. However, calculating the right hand side of the formula above comes out as $2 \lfloor t \rfloor( t - \lfloor t \rfloor)$, which is not always $0$. There could well be an error in my calculation, and certainly the formula you give holds in the case of a Poisson process. – owen88 Mar 19 '15 at 10:06
• In light of my answer posted below, clearly there is a problem with my counter example... Or if may be that the claim I state below fails in the case that the function in the convolution is a step function. – owen88 Mar 19 '15 at 11:25

I too was unable to derive a direct proof of the formula you have given; the following proof therefore is somewhat indirect. It rests on the following claim.

Claim. Given functions $H(t), \, f(t)$, the renewal equation $g(t) = H(t) + \int_{0}^t g(t-x) f(x) \mathrm d x$ has a unique solution.

Henceforth I will use the normal notation for convolution $\phi*\psi(t) = \int_0^t \phi(t-x)\psi(x) \mathrm d x$ so simplify notation. The equation above then reads: $g = H + g*f$.

My proof proceeds in two steps:

1. Derive a renewal type equation for the variance.
2. Prove that it is solved by the right hand side of the formula you gave.

Part 1 Instead of working with the variance, i will consider $g(t) = \textbf{E}[N(t)^2]$. Exactly as though we were deriving the usual renewal equation, we condition on the first jump time \begin{align*} g(t) & = \int_{0}^\infty \textbf{E}[N(t)^2 \, | \, X_1 = x] f(x) \mathrm d x \\ & = \int_0^t \textbf{E}[N(t)^2 \, | \, X_1 = x] f(x) \mathrm d x + \int_t^\infty 0 \mathrm \, d x \\ & = \int_0^t \textbf{E}[(1 + N(t-x) )^2] f(x) \mathrm d x \\ & = \int_0^t f(x)\Big(1 + 2 \textbf{E}[N(t-x)] + \textbf{E}[N(t-x)^2] \Big ) \mathrm d x \\ & = F(t) + 2(m * f)(t) + (g*f)(t). \end{align*} In particular, $g$ satisfies a renewal type equation with $H = F +2 m*f$.

Part 2 According to the formula you gave, we want to show $g(t) = 2(m*m')(t) + m(t)$, note that this is exactly the right hand side of your formula but with the $m(t)^2$ term removed, since we are not working with the variance. Substituting this formula into the right hand side of the renewal equation we have \begin{align} H + (2(m*m') + m)*f & = F + 2(m*f) +2(m*m')*f + m*f\\ & = (m - m*f) + 2(m*f) +2(m*m')*f + m*f\\ & = m + 2( m*f + m*m'*f), \end{align} where in the second line we substituted in the renewal equation $F = m - m*f$. It remeains then to show \begin{align} m*f + m*m'*f =m*m'. \tag{1} \end{align} But again using the renewal equation, and differentiating it \begin{align} m' = (F + m*f)' = f +m'*f, \end{align} which when rearranged is $$m'*f =m'-f$$ Substituting into (1) \begin{align} m*f + m*m'*f & = m*f + m*(m' -f) \\ & =m*m', \end{align} as required.