How do we know whether $X_{i}$'s are independent and what is the specific nature of $f$ given in the integral?

Question

The problem is given as :-

For a continuous real valued function $\Large f$, defined on $\Large [0,1]$ calculate :- $\Large \underset{n \to \infty}{\lim} \underset{0}{\overset{1}{\int}} \cdots \underset{0}{\overset{1}{\int}} f \Bigg((x_{1} \cdots x_{n})^{\frac1n} \Bigg) dx_{1} \cdots dx_{n}$

I got the idea from here and here about how to proceed to evaluate this integral but got stuck too at several parts of the problem. So here is my approach.

It is clear that $f$ is bounded and continuous and we need to find $\large \underset{n \to \infty}{\lim}\mathbf{\mathbb{E}}\Bigg[f \Bigg((\underset{i=1}{\overset{n}{\Pi}} x_{i})^{\frac1n}\Bigg) \Bigg]$ and $X_{i} \sim U[0,1]$

Now consider $y= \underset{i=1}{\overset{n}{\Pi}}(x_{i})^{\frac1n}\\ \implies \log y= \frac1n \underset{i=1}{\overset{n}{\sum}} \log x_{i}\\ \implies -\log y= - \frac1n \underset{i=1}{\overset{n}{\sum}} \log x_{i} \qquad \longrightarrow (i)$

$X_{i}$'s are identical since $X_{i} \sim U[0,1], \qquad \forall i \in \{1,2, \cdots, n \}$

then each of $-\log X_{i} \sim \exp(1)$

But here is my first doubt:-

$1•$ I'm not sure whether each of $X_{i}$ are independent of each other.

If they are then $ \frac1n \underset{i=1}{\overset{n}{\sum}} - \log x_{i} \longrightarrow 1 \qquad \textit{ by slln. } \\ \implies \frac1n \underset{i=1}{\overset{n}{\sum}} \log x_{i} \longrightarrow -1 \\ \implies (\underset{i=1}{\overset{n}{\Pi}} x_{i})^{\frac1n} \longrightarrow e^{-1} \\ \implies f \Bigg((\underset{i=1}{\overset{n}{\Pi}} x_{i})^{\frac1n}\Bigg) \longrightarrow f(e^{-1}) \\ \implies \mathbf{\mathbb{E}}\Bigg[f \Bigg((\underset{i=1}{\overset{n}{\Pi}} x_{i})^{\frac1n}\Bigg) \Bigg] \longrightarrow \mathbf{\mathbb{E}}\Bigg[f(e^{-1}) \Bigg] \\ \implies \underset{n \to \infty}{\lim}\mathbf{\mathbb{E}}\Bigg[f \Bigg((\underset{i=1}{\overset{n}{\Pi}} x_{i})^{\frac1n}\Bigg) \Bigg] \longrightarrow \underset{n \to \infty}{\lim}\mathbf{\mathbb{E}}\Bigg[f(e^{-1}) \Bigg] \longrightarrow \underset{\text{by dct}}{\underbrace{\mathbb{E}(\underset{n \to \infty}{\lim}f(e^{-1}))}} \longrightarrow \mathbb{E}\underset{\text{since } f \\ \text{ is continuous and bounded}}{\underbrace{(f(\underset{n \to \infty}{\lim}e^{-1}))}} \longrightarrow \mathbf{\mathbb{E}}\Bigg[f(e^{-1}) \Bigg] \\ \implies \underset{n \to \infty}{\lim}\mathbf{\mathbb{E}}\Bigg[f \Bigg((\underset{i=1}{\overset{n}{\Pi}} x_{i})^{\frac1n}\Bigg) \Bigg] \longrightarrow \mathbf{\mathbb{E}}\Bigg[f(e^{-1}) \Bigg] \longrightarrow f(e^{-1}) \qquad \to (ii)$

Now if $f(e^{-1})$ has any specific value $c$ then $\underset{n \to \infty}{\lim}\mathbf{\mathbb{E}}(c) = c$ as we can see in this problem, where the specific nature of $f$ has been given.

So from $(ii)$ my question is :-

$2•$ What is the value of $f(e^{-1})$ ?
$3•$ Also, can we say that a continuous and bounded function of continuous uniform random variable, over the region $[0,1]$ will also be uniform over $[0,1]$?

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My second approach

Also from $(i)$ we have,

$- \underset{i=1}{\overset{n}{\sum}} \log X_{i} \sim \gamma(n, 1) $ and $- \frac1n \underset{i=1}{\overset{n}{\sum}} \log X_{i} \sim \gamma(n,n)$

$-\log y= - \frac1n \underset{i=1}{\overset{n}{\sum}} \log x_{i} \\ \implies e^{-\log y}= e^{- \frac1n \underset{i=1}{\overset{n}{\sum}} \log x_{i}} \\ \implies \frac1y = e^{\mathbb{E}{[- \log x_{i}]}} \qquad [ \because - \frac1n \underset{i=1}{\overset{n}{\sum}} \log x_{i} = \mathbb{E}{[- \log x_{i}]} \\ \implies \frac1y = e^{n} \qquad [\because \mathbb{E}{[- \log x_{i}]} = n , \forall i \in \{1,2, \cdots, n\}] \\ \implies y = e^{-n} \\ \implies f(y) = f(e^{-n}) \\ \implies \mathbb{E}(f(y)) = \mathbf{\mathbb{E}}\Bigg[f \Bigg((\underset{i=1}{\overset{n}{\Pi}} x_{i})^{\frac1n}\Bigg) \Bigg] = \mathbb{E}(f(e^{-n})) \qquad \longrightarrow (iii)$

From $(iii)$, here are my other doubts :-

$4•$ What is the value of $f(e^{-n})$ here?
$5•$ Can we say that $\underset{n \to \infty}{\lim} \mathbb{E}(f(y)) = \underset{n \to \infty}{\lim} \mathbf{\mathbb{E}}\Bigg[f \Bigg((\underset{i=1}{\overset{n}{\Pi}} x_{i})^{\frac1n}\Bigg) \Bigg] = \underset{n \to \infty}{\lim} \mathbb{E}(f(e^{-n})) = \underset{\text{by dct}}{\underbrace{\mathbb{E}(\underset{n \to \infty}{\lim}f(e^{-n}))}} = \mathbb{E}\underset{\text{since } f \\ \text{ is continuous and bounded}}{\underbrace{(f(\underset{n \to \infty}{\lim}e^{-n}))}}= \mathbb{E}(f(0))?$
I am not sure what $\mathbb{E}(f(0))$ would be.

But is this approach correct?

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Alternatively if we write the integral for $n=1,2, \cdots$ respectively, then we see that the $f$ is continuous, bounded and (looks like) monotonically decreasing (hence bijective). But does it help in evaluating the integral ?

Hence any help or explanation in this regard will be helpful, valuable and much appreciated.

Answer 1

The short answer: We will know that the $X_i$ are independent because we are the ones introducing the $X_i$, so we will be in a position to declare that they are independent. We don't know the specific nature of $f$. The result remains true for a general $f$, so the final result will have to be left in terms of $f$.

1. When placing the integral in the context of expectation, you are the one who gets to (or, in reality, has to) specify the joint distribution. Specifying $X_i\sim \text{Uniform}(0,1)$ isn't sufficient to determine independence, because independence has to do with the joint rather than the marginal distributions. For example, assume $U,V$ are independent $\text{Uniform}(0,1)$. Then the random vector $X=(U,V)$ has the same marginals as $Y=(U,U)$, but the coordinates of $X$ are independent, while the coordinates of $Y$ are not. Checking independence in the case of continuous random variables is the same as checking whether the joint pdf splits into the product of the marginal pdfs ($\color{red}{\text{caution}}$). In this case, we want to get the joint pdf to be uniform on the cube $[0,1]^n$, so the product of the uniform pdfs on the segments $[0,1]$ is exactly what we want. So you have to specify independence, which you have the ability to do, since you are the one introducing the $X_i$.

$\color{red}{\text{caution}}$ This can be a little bit tricky. When I talk about pdfs, I always assume they're defined on all of $\mathbb{R}$. However, many times and many people have the convention of only defining the pdf on its support (ie for exponential, they may say $f(x)=e^{-x/\lambda}/\lambda$, $x>0$, and not specify any definition for $x\leqslant 0$). So when I say the join pdf splits into the product of the marginals, I mean they split as functions defined on all of $\mathbb{R}$. However, if you only define the joint pdf on its support, it may appear that it splits, when actually the "non-splitting" sneaks in through the domain. For example, say $$f(x,y)=\left\{\begin{array}{ll} cxy & : 0\leqslant x\leqslant x+y\leqslant 1 \\ 0 & : \text{otherwise.}\end{array}\right.$$ It looks like it splits, but if you try to write this function as $h(x)g(y)$ where $h,g$ are each defined on all of $\mathbb{R}$, you can't do it because of the $0\leqslant x\leqslant x+y\leqslant 1$ condition.

2. We can't know the value of $f(e^{-1})$ without knowing $f$. Since the purpose of this problem seems to be to deal with a general $f$, you can't be more specific than $f(e^{-1})$.

3. If I understand your question, you're asking whether $f(U)$ will be $\text{Uniform}(0,1)$ if $U$ is $\text{Uniform}(0,1)$ and $f:[0,1]\to \mathbb{R}$ is continuous (and therefore bounded). The answer is no. Take for example $f(U)=2U$, which is $\text{Uniform}(0,2)$, or $f(U)=U^2$.

4. Again, we can't be more specific about the value of $f(e^{-n})$.

5. Yes. Your steps are exactly correct here.

There is one step in your second approach (I didn't check whether it occurs once or more than once throughout your post) where you have written $$f\Bigl(\prod_{i=1}^n (x_i)^{\frac{1}{n}}\Bigr) = \prod_{i=1}^n f(x_i)^{\frac{1}{n}}.$$ This occurs in the line before "From $(iii)$, here are my other doubts." On the left, the exponent $1/n$ is inside of $f$ (which is correct), and on the right, it's on the outside (which is not correct). In general, these need not be the same. It looks like you know this, because after this you write $f(e^{-n})$, which is correct again. So it looks to me like you wrote an intermediate step that you didn't actually mean. In any case, it's a small issue and didn't impact the final result.