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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 $JK$?

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  • $\begingroup$ What are $g$ and $\delta_t$? $\endgroup$
    – user39119
    Dec 13, 2019 at 18:18
  • $\begingroup$ g is any function you like. delta t is a discrete change in, say, time (but doesn't matter what it represents, really). $\endgroup$
    – buckner
    Dec 13, 2019 at 18:46
  • $\begingroup$ 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 $\delta$. $\endgroup$
    – Attack68
    Dec 13, 2019 at 19:46
  • $\begingroup$ Independent is easy. The question is whether it is true whether independent or not. And clearly X and Y are dependent on delta t. $\endgroup$
    – buckner
    Dec 13, 2019 at 19:50

2 Answers 2

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

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A discrete distribution solution

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

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