I'm taking a mathematical statistics course and I'm unsure of one of the steps in a homework problem. The problem is to find the plug-in estimator for the parameter $p$ in a $t$-step random walk, given iid sample $X_i, ..., X_n$.
Definition:
If $T(F)=\int{r(x)dF(x)}$ for some function $r(x)$, then $T$ is a linear functional. ($F$ is cdf of the related random variable.)
We're given:
$$\ X=\sum_{i=1}^tY_i,\qquad Y_i = \begin{cases} -1, & \text{w.p. 1-p} \\ +1 , & \text{w.p. p } \end{cases}$$
and
$$E[X] = t(1-2p)$$ (and therefore) $$p=\frac{1}{2}(1-\frac{E[X]}{t})$$
I have a theorem in my book that gives the plug in estimator, assuming the parameter of interest is a linear functional.
I believe from the definition that $p$ is not a linear functional, since it's not expressed in the required form, but I'm not sure that there isn't some way to do so, and therefore I don't know whether the theorem applies.
So the question is: How do I determine for sure whether p is a linear functional?
Linear in what? From your notation I suspect you ask whether $p$ is a linear functional on the space of functions of bounded variation $BV$. First note that \begin{align*} p(F) &= \frac 12 \left(1 - \frac 1t \int x\,\mathrm dF(x)\right) \\ &= \frac 12 \left(\int\,\mathrm dF(x) - \frac 1t \int x\,\mathrm dF(x)\right) \end{align*} since $$\int\,\mathrm dF(x) = F(\infty) - F(-\infty) = 1 - 0 = 1.$$ Let $F,G\in BV$, and $c\in\mathbb R$. Then \begin{align*} p(cF + G) &= \frac 12 \left(\int\,\mathrm d(cF(x) + G(x)) - \frac 1t \int x\,\mathrm d(cF(x) + G(x))\right) \\ &= \frac 12 \left(c\int\,\mathrm dF(x) + \int\,\mathrm dG(x) - \frac ct \int x\,\mathrm dF(x) - \frac 1t \int x\,\mathrm dG(x)\right) \\ &= \frac 12 \left(c\int\,\mathrm dF(x) - \frac ct\int x\,\mathrm dF(x) + \int\,\mathrm dG(x) + \frac 1t\int x\,\mathrm dG(x)\right) \\ &= c\left(\frac 12\left( \int\,\mathrm dF(x) - \frac 1t\int x\,\mathrm dF(x)\right)\right) + \frac 12\left(\int\,\mathrm dG(x) + \frac 1t\int x\,\mathrm dG(x)\right) \\ &= cp(F) + p(G).\end{align*} That is, $p$ is a linear functional.
Edit: There was a mistake in my original answer. I was tempted to say that $p$ is a linear a functional on $BV$. However, we have that $H\in BV$, where $H(x) = 0$ for all $x$. Obviously, the equality $$\int\,\mathrm dH(x) = 1,$$ which is crucial for the proof, cannot hold then. Hence $p$ cannot be a linear functional on all of $BV$. It is only true that $p$ is linear for a particular subset (namely the set of all functions satisfying the above equality). This subset is, however, not a linear subspace. Hence the terminology used in the reference sounds somewhat ambigous to me...