How to model a stochastic process, continuous in stepsize, which converges against a simple random walk? I want to compute the probability distribution for a stochastic process with discrete number of steps, where each real value has a nonvanishing probability to be the next stepsize. And I want to introduce a parameter $t$ such that I end up with a simple finite stepsize random walk in the limit of small/big $t$.
For a random walk with steps $+1$ and $-1$ with probabilities $p_+$ and $p_-=1-p_+$, the characteristic function is
$$\phi(a)=\sum_{k=1}^2p_k \mathrm e^{ikx}$$
and when I add up $n$ steps, I end up with 
$$(p_-\mathrm e^{ix}+p_+\mathrm e^{-ix})^n$$ 
which must be fourier transformed. I draw my information from this book on Kinetic theory, but that one is about physics, really.I also find these terms on the Wikipedia page Binomial distribution with somewhat different notation. 
Now my attempt for the continuous approach is to replace "$+1$ and $-1$ with probabilities $p_+$ and $p_-=1-p_+$" by the continuous distribution with two peaks, i.e. if $\lambda\in\mathbb R$ is the stepsite, the probability should be, I think 
$$a_+(t)\ \mathrm e^{-(\lambda-1)^2/2t}+a_-(t)\ \mathrm e^{-(\lambda+1)^2/2t}$$
with the right normalizations $a_\pm(t)$. But how to compute that to the power of the stepsize $(\dots)^n$ to obtain the probability? 
If the integral is too difficult, can we approximately solve it? Or is there a more reasonable approach? For starters, $p_+=p_-=\tfrac{1}{2}$ is of interest. One could of course also make variable peaks, which at the moment are set to $+1$ / $-1$. 
 A: This question seems to revolve around the idea of mollifying the $\pm1$ steps of a discrete random walk by some gaussian distributions of smaller and smaller width. 
In terms of random variables, one replaces each $\pm1$ step $X$ such that $P[X=+1]=p$ and $P[X=-1]=q$ with $q=1-p$, by a step $X^t$ defined by
$$
X^t=X+Z^t,
$$ 
where $Z^t$ is normal $N(0,t)$. The density $f^t$ of the distribution of $X^t$ is such that
$$
f^t(x)=\frac{p\mathrm e^{-(x-1)^2/2t}+q\mathrm e^{-(x+1)^2/2t}}{\sqrt{2\pi t}}.
$$
Likewise, the sum of $n\geqslant1$ i.i.d. steps distributed like $X^t$ is 
$$
S^t_n=S_n+Z^t_n,
$$ 
where $S_n$ is the sum of the $\pm1$ steps $X$, hence $\frac12(S_n+n)$ is binomial $(n,p)$, and $Z^t_n$ is the sum of the normal steps $Z^t$ hence $Z^t_n$ is normal $N(0,nt)$. The density $f_n^t$ of the distribution of $S^t_n$ is the convolution of $f^t$ with itself $n$ times, that is, $f_n^t$ is such that
$$
f_n(x)=\frac1{\sqrt{2\pi nt}}\sum_{k=0}^n{n\choose k}p^kq^{n-k}\mathrm e^{-(x-2k+n)^2/2nt}.
$$
To realize every random variable $S_n^t$ on a common probability space (hence being able to consider almost sure limits), simply consider
$$
S^t_n=\sum_{k=1}^nX_k+\sqrt{t}\sum_{k=1}^nY_k,
$$
where the sequence $(X_k)_{k\geqslant1}$ is i.i.d. $\pm1$ Bernoulli, the sequence $(Y_k)_{k\geqslant1}$ is i.i.d. standard normal and the two sequences are independent.
