As already mentioned by Shai, this is standard renewal theory. In the limit $K\to+\infty$, the excess $Y_\tau-K$ converges in distribution (for a non asymptotic result, go to the very end of this post). The most appealing (to me) description of this convergence result is as follows.

First the length $X_{\tau}$ of the renewal interval $[Y_{\tau-1},Y_\tau[$ which contains $K$ converges in distribution to the size-biased distribution of the holding times $X$. In other words, $X_\tau$ converges in distribution to a random variable $\hat{X}$ whose distribution is characterized by the fact that $E(u(\hat{X}))=E(Xu(X))/E(X)$ for every bounded measurable function $u$. Second the location of $K$ in the renewal interval $[Y_{\tau-1},Y_\tau[$ is uniformly distributed.

This shows that the overshoot $Y_\tau-K$ converges in distribution to $U\hat{X}$ where $\hat{X}$ is as above, $U$ is uniform on $[0,1]$ and $\hat{X}$ and $U$ are independent. In particular, $$ E(Y_\tau-K)\to E(U\hat{X})=E(X^2)/(2E(X)), $$ in the usual sense if $X$ is square integrable and in the sense that it converges to $+\infty$ if $X$ is not square integrable.

At least, this is the situation when the holding times $X$ are continuous random variables. Now, I realize that the OP is interested in integer valued holding times, in which case one should assume that $X$ is not restricted to a sublattice of the integers and one should modify the result as follows. The renewal length $X_\tau$ still converges in distribution to $\hat{X}$ defined as before (and in the discrete case the distribution of $\hat{X}$ is simply given by $P(\hat{X}=k)=kP(X=k)/E(X)$ for every $k$). But now the overshoot $Y_\tau-K$ converges in distribution to a random variable $Z$ which is uniformly distributed on the set $\{1,2,\ldots,\hat{X}\}$ conditionally on $\hat{X}$. For instance, $$ E(Y_\tau-K)\to E(Z)=E(\hat{X}+1)/2=E(X(X+1))/(2E(X)). $$ Finally, a non asymptotic upper bound of the distribution of the overshoot $Y_\tau -K$ is $$ P(Y_\tau -K=n)\le P(X\ge n), $$ for every $n$ and every $K$. This implies that, for every $K$, $$ E(Y_\tau -K)\le E(X(X+1))/2. $$

As mentioned in Shai's answer, this is standard renewal theory. In the limit $K\to+\infty$, the overshoot $Y_\tau-K$ converges in distribution. (For a non asymptotic result, go to the very end of this post.) The most appealing (to me) description of this convergence result is as follows.

First the length $X_{\tau}=Y_{\tau}-Y_{\tau-1}$ of the renewal interval $[Y_{\tau-1},Y_\tau[$ which contains $K$ converges in distribution to the size-biased distribution of the holding times $X$. This means that $X_\tau$ converges in distribution to a random variable $\hat{X}$ whose distribution is characterized by the fact that, for every bounded measurable function $u$, $$ E(u(\hat{X}))=\frac{E(Xu(X))}{E(X)}. $$ Second the location of $K$ in the renewal interval $[Y_{\tau-1},Y_\tau[$ is uniformly distributed.

This shows that the overshoot $Y_\tau-K$ converges in distribution to $U\hat{X}$ where $\hat{X}$ is as above, $U$ is uniform on $[0,1]$ and $\hat{X}$ and $U$ are independent. In particular, $$ \lim_{K\to+\infty}E(Y_\tau-K)=E(U\hat{X})=E(U)E(\hat{X})=\frac{E(X^2)}{2E(X)}, $$ in the usual sense if $X$ is square integrable and in the sense that the limit is $+\infty$ if $X$ is not square integrable.

At least, this is the situation when the holding times $X$ are continuous random variables. Now, I realize that the OP is interested in integer valued holding times, in which case one should assume that $X$ is not restricted to a sublattice of the integers and one should modify the result as follows. 

The renewal length $X_\tau$ still converges in distribution to $\hat{X}$ defined as before. (And in the discrete case the distribution of $\hat{X}$ is simply given by $P(\hat{X}=k)=kP(X=k)/E(X)$ for every $k$.) But now the overshoot $Y_\tau-K$ converges in distribution to a random variable $Z$ which is uniformly distributed on the set $\{1,2,\ldots,\hat{X}\}$ conditionally on $\hat{X}$. For instance, $$ \lim_{K\to+\infty}E(Y_\tau-K)=E(Z)=\frac12E(\hat{X}+1)=\frac{E(X(X+1))}{2E(X)}. $$ Finally, a non asymptotic upper bound of the distribution of the overshoot $Y_\tau -K$ is the fact that $$ P(Y_\tau -K=n)\le P(X\ge n), $$ for every $n$ and every $K$. This implies that, for every $K$, $$ E(Y_\tau -K)\le\frac12E(X(X+1)). $$

As already mentioned by Shai, this is standard renewal theory. In the limit $K\to+\infty$, the excess $Y_\tau-K$ converges in distribution. The most appealing (to me) description of this result is as follows.

First the length $X_{\tau}$ of the renewal interval $[Y_{\tau-1},Y_\tau]$ which contains $K$ converges in distribution to the size-biased distribution of the holding times $X$. In words, $X_\tau$ converges in distribution to a random variable $\hat{X}$ whose distribution is characterized by the fact that $E(u(\hat{X}))=E(Xu(X))/E(X)$ for every bounded measurable function $u$. Second the location of $K$ in the renewal interval $[Y_{\tau-1},Y_\tau]$ is uniformly distributed.

This shows that the excess $Y_\tau-K$ converges in distribution to $U\hat{X}$ where $\hat{X}$ is as above, $U$ is uniform on $[0,1]$ and $\hat{X}$ and $U$ are independent. In particular, $$ E(Y_\tau-K)\to E(U\hat{X})=E(X^2)/(2E(X)), $$ in the usual sense if $X$ is square integrable and in the sense that it converges to $+\infty$ if $X$ is not square integrable.

At least, this is the situation for continuous random variables $X$. Now, I realize that the OP is interested in integer random variables, in which case one should assume that $X$ is not restricted to a sublattice of the integers and one should modify the result as follows. The renewal length $X_\tau$ converges in distribution to $\hat{X}$ defined as before (and in this discrete case the distribution of $\hat{X}$ is simply given by $P(\hat{X}=k)=kP(X=k)/E(X)$ for every $k$). But now $Y_\tau-K$ converges in distribution to a random variable $Z$ uniform on the set $\{1,2,\ldots,\hat{X}\}$ conditionally on $\hat{X}$. For instance, $$ E(Y_\tau-K)\to E(Z)=E(\hat{X}+1)/2=E(X(X+1))/(2E(X)). $$

As already mentioned by Shai, this is standard renewal theory. In the limit $K\to+\infty$, the excess $Y_\tau-K$ converges in distribution (for a non asymptotic result, go to the very end of this post). The most appealing (to me) description of this convergence result is as follows.

First the length $X_{\tau}$ of the renewal interval $[Y_{\tau-1},Y_\tau[$ which contains $K$ converges in distribution to the size-biased distribution of the holding times $X$. In other words, $X_\tau$ converges in distribution to a random variable $\hat{X}$ whose distribution is characterized by the fact that $E(u(\hat{X}))=E(Xu(X))/E(X)$ for every bounded measurable function $u$. Second the location of $K$ in the renewal interval $[Y_{\tau-1},Y_\tau[$ is uniformly distributed.

This shows that the overshoot $Y_\tau-K$ converges in distribution to $U\hat{X}$ where $\hat{X}$ is as above, $U$ is uniform on $[0,1]$ and $\hat{X}$ and $U$ are independent. In particular, $$ E(Y_\tau-K)\to E(U\hat{X})=E(X^2)/(2E(X)), $$ in the usual sense if $X$ is square integrable and in the sense that it converges to $+\infty$ if $X$ is not square integrable.

At least, this is the situation when the holding times $X$ are continuous random variables. Now, I realize that the OP is interested in integer valued holding times, in which case one should assume that $X$ is not restricted to a sublattice of the integers and one should modify the result as follows. The renewal length $X_\tau$ still converges in distribution to $\hat{X}$ defined as before (and in the discrete case the distribution of $\hat{X}$ is simply given by $P(\hat{X}=k)=kP(X=k)/E(X)$ for every $k$). But now the overshoot $Y_\tau-K$ converges in distribution to a random variable $Z$ which is uniformly distributed on the set $\{1,2,\ldots,\hat{X}\}$ conditionally on $\hat{X}$. For instance, $$ E(Y_\tau-K)\to E(Z)=E(\hat{X}+1)/2=E(X(X+1))/(2E(X)). $$ Finally, a non asymptotic upper bound of the distribution of the overshoot $Y_\tau -K$ is $$ P(Y_\tau -K=n)\le P(X\ge n), $$ for every $n$ and every $K$. This implies that, for every $K$, $$ E(Y_\tau -K)\le E(X(X+1))/2. $$