# Limit of product

Suppose $$g(X, \delta_t)$$ approaches a constant $$J$$ as $$\delta_t$$ approaches $$0$$, where $$X$$ is a random variable, and suppose $$Y^2/\delta_t$$ approaches some constant $$K$$ as $$\delta_t$$ approaches $$0$$, where $$Y$$ is also random.

Then how can we prove that $$E[g(X, \delta_t)\times Y^2/\delta_t]$$ approaches $$JK$$?

• What are $g$ and $\delta_t$?
– user39119
Dec 13 '19 at 18:18
• g is any function you like. delta t is a discrete change in, say, time (but doesn't matter what it represents, really). Dec 13 '19 at 18:46
• If X and Y are are independent, then the expectation is separable and the result immediately follows. So are you interested in the case they are dependent? Additionally X and Y must be dependent upon $\delta$.
– Attack68
Dec 13 '19 at 19:46
• Independent is easy. The question is whether it is true whether independent or not. And clearly X and Y are dependent on delta t. Dec 13 '19 at 19:50

I think you should be able to show this by expanding your probability density function and showing it collapses to a dirac delta function:

e.g.

For $$Y^2/\delta_t \rightarrow K$$ as $$\delta_t \rightarrow 0$$, then clearly $$E[Y^2/\delta_t] \rightarrow K$$ also.

So $$\int_Y y^2/\delta_t p(y, \delta_t) .dy = \int_Yy^2/\delta_t [ p(y,0) + \delta_t p'(y,0) +..].dy \rightarrow K$$

Implies that $$p(y, \delta_t) \rightarrow \delta_t dirac(\sqrt{K}-y)$$

Similarly $$p_X(x,\delta_t) = dirac(g^{-1}(J,0) - x)$$

and the 2D version $$p_{XY}(x,y,\delta_t) = \delta_t dirac^2(g^{-1}(J,0) - x, \sqrt{K} - y) = dirac(g^{-1}(J,0) - x)\delta_tdirac(\sqrt{K} - y)$$

So $$\lim_{\delta_t \rightarrow 0 } E[g(X,\delta_t) Y^2/\delta_t] = \int_X g(x,0) dirac(g^{-1}(J,0)-x).dx \int_Y (y^2/\delta_t) \delta_t dirac(\sqrt{K}-y).dy = JK$$

Could the following proof do for the above. I have done only for a discrete distribution but can be expanded to continuous