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Let $Y_n$ be a sequence of random variable such that $$ \sqrt{n}(Y_n-\mu) \stackrel{d}{\to} \mathcal{N}(0, \sigma^2), $$ and thus we can say $Y_n$ is asymptotically normally distributed as $$ Y_n \stackrel{a}{\sim} \mathcal{N}\bigg(\mu, \frac{\sigma^2}{n}\bigg). $$ Now suppose want to approximate $E[f(Y_n)]$. By a Taylor series expansion we have $$ E[f(Y_n)] \approx f(E[Y_n]) + \frac{f''(E[Y_n])}{2}\text{Var}(Y_n). $$

It seems that this means that as $n\to \infty$ we can make use of the asymptotic distribution of $Y_n$, i.e., we are allowed to say: $$ E[f(Y_n)] \approx f(\mu) + \frac{f''(\mu)}{2}\frac{\sigma^2}{n}, \quad \quad \text{as} \ n \to \infty. $$

Does this expression hold? Do we need some assumptions before we can say it? One of the reasons I am not certain is that I read here that convergence in distribution just means the CDFs of the random variable is becoming closer to the limit CDF, but the actual values of the random variable may not be becoming closer together to the values of the limiting random variable. We need convergence in probability for the values to become closer.

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    $\begingroup$ The symbol $\approx$ being vague, what do you mean by it [in a mathematical sense] ? $\endgroup$
    – Xi'an
    Commented Nov 6, 2020 at 14:26
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    $\begingroup$ Even when a sequence of random variables converges in distribution, the corresponding sequence of expectations needn't converge at all. A standard example is the sequence $X_n=nY_n,$ $n=1,2,3,\ldots,$ where $Y_n$ has a Bernoulli$(p(n))$ distribution and $p(n)$ is chosen to converge to $0$ (so that $X_n$ converges to $0$ in distribution) in such a way that $E[X_n]=np(n)$ does not converge; e.g., $p(n)=1/\sqrt{n}.$ $\endgroup$
    – whuber ♦
    Commented Nov 6, 2020 at 14:35
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    $\begingroup$ @whuber In general they need not converge, but my situation is more specific. I have a sequence of random variables that are asymptotically normal. Maybe in my case the convergence in distribution can pass over to the convergence in expectation? $\endgroup$
    – Bertus101
    Commented Nov 6, 2020 at 15:09
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    $\begingroup$ It doesn't work that way: for instance, we could start with a sequence of random variables like yours whose expectations do converge and add my sequence to them. The new sequence is still asymptotically Normal but its expectation diverges. $\endgroup$
    – whuber ♦
    Commented Nov 6, 2020 at 15:47
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    $\begingroup$ That would work provided you use the absolute value of $f$, as required by that theorem. The Taylor series approach does not necessarily work: you need to adduce additional conditions to justify omitting the remainder term in the Taylor expansion and you need to assume that sufficiently high moments of $Y_n$ are bounded. $\endgroup$
    – whuber ♦
    Commented Nov 17, 2020 at 17:10

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