# Show that $Y_n := (\prod_{i=1}^{n} X_i)^{1/n}$ converges with probability 1

I'm dealing with a problem about stochastic and statistics and hope some of you can help me!

On $$[0,1]$$ we have a sequence of independent, equally distributed probability variables $$(X_n)_{n \in \mathbb{N}}$$. I have to show:

a) $$Y_n := (\prod_{i=1}^{n} X_i)^{1/n}$$ converges with probability $$1$$.

b) calculate the exact limit of $$Y_n$$

I've already done some calculations, but I'm really not sure, whether everything is fine.

Some pre-considerations: To get rid of the product I took the logarithm: $$\ln(Y_n) = \frac{1}{n} \sum_{i=1}^{n} \ln(X_i)$$

After taking the logarithm the sequence $$\ln(X_i)$$ still is equally distributed and independent.

a) I found a theorem in my lecture notes, which states, that $$\frac{S_n}{n}$$ (the $$n$$-th partial sum of a sequence) converges and has a finite limit with probability $$1$$ if the sequence is integrable.

It seems to me, that this Theorem might fit, but my concern is, that the logarithm of $$X_i = 0$$ (allowed since $$X_i$$ is a sequence on $$[0,1]$$) isn't integrable.

b) This part some kind of "smells" to me like Kolmogorov's law, which states that for a sequence of indempendent and identically distributed probability variables with finite expectation value it holds: $$\lim_{n\rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n}X_k = \mathbb{E}(X_1) \quad \text{a.s.}$$

So the limit would be $$\lim_{n} \ln(Y_n) = \mathbb{E}(\ln(X_1))$$ almost sure.

But I don't see, why the expectation value of $$\ln(X_i)$$ should be finite for $$X_i = 0$$ again.

So due to this concerns at $$X_i = 0$$ I'm not sure, whether I'm on the right track, or the problem needs to be solved differently.

I would be very grateful if some of you can help me!

pcalc

• Note that if one of the $X_i$ is $0$ the whole product is $0$ thus $$Y_n=\left( \prod_{i=1}^n X_i 1_{X_i\in (0,1]}\right)^{1/n}$$ and the $X_i 1_{X_i\in (0,1]}$ are still i.i.d so the SLLN still applies (provided $\log X$ is integrable). – Gabriel Romon Feb 1 '19 at 15:56
• Hi and thanks for your quick response! So - if I take this special case for $X_i=0$ in concern - my way is suitable? – pcalc Feb 1 '19 at 16:19
• @GabrielRomon As far as I understand, the OP's main problem is that there is no assumption that $\log(X 1_{\{X>0\}})$ is integrable. – saz Feb 1 '19 at 17:07

If $$\mathbb{E}(-\log(X_1))<\infty$$ then your reasoning works fine and we find that

$$Y_n \to \exp(\mathbb{E}\log(X_1)) \quad \text{almost surely}. \tag{1}$$

Now consider the case $$\mathbb{E}(-\log(X_1))=\infty$$. Define a sequence of truncated random variables by

$$Z_n^{(k)} := \min\{k, -\log(X_n)\}= \begin{cases} - \log(X_n), & 0 \leq -\log(X_n) \leq k, \\ k, & \text{otherwise}. \end{cases}$$

The sequence $$(Z_n^{(k)})_{n \in \mathbb{N}}$$ is independent and identically distributed. Since $$\mathbb{E}|Z_n^{(k)}| \leq k < \infty$$, the strong law of large numbers gives

$$\lim_{n \to \infty} \frac{1}{n} \sum_{j=1}^n Z_j^{(k)} \xrightarrow[]{n \to \infty} \mathbb{E}(Z_1^{(k)}) \tag{1}$$

almost surely. Since $$Z_j^{(k)} \leq - \log(X_j)$$ for each $$j \in \mathbb{N}$$ this implies

$$\liminf_{n \to \infty}\frac{1}{n} \sum_{j=1}^n -\log(X_j) \geq \mathbb{E}(Z_1^{(k)})$$

for all $$k \in \mathbb{N}$$. Since the monotone convergence theorem gives $$\sup_k \mathbb{E}(Z_1^{(k)}) = \mathbb{E}(-\log(X_1))=\infty$$ we get

$$\liminf_{n \to \infty} \frac{1}{n} \sum_{j=1}^n -\log(X_j) = \infty$$ i.e.

$$\limsup_{n \to \infty} \frac{1}{n} \sum_{j=1}^n \log(X_j) = -\infty$$

almost surely. Hence, by the continuity of the exponential function,

$$Y_n = \exp\left( \frac{1}{n} \sum_{j=1}^n \log(X_j) \right) \xrightarrow[]{n \to \infty} 0$$

almost surely.

In summary, we get

$$Y_n \to \exp(\mathbb{E}\log(X_1)) \quad \text{a.s.}$$

with $$\mathbb{E}\log(X_1)$$ being possibly $$-\infty$$.

Remark: We have actually proved the following converse of the strong law of large numbers:

Let $$(U_j)_{j \in \mathbb{N}}$$ be a sequence of independent identically distributed and non-negative random variables. If $$\mathbb{E}(U_1)=\infty$$ then $$\liminf_{n \to \infty} \frac{1}{n} \sum_{j=1}^n U_j = \mathbb{E}(U_1)=\infty \quad \text{a.s.}.$$

• Hey Saz, I see that there is no need to split cases like I did "events such that of $X_n$ is zero.." . I'll delete my answer since yours provides a good neat answer. (+1) – Shashi Feb 1 '19 at 17:55
• Hi! Thank you very much for your very clear answer! That helped me a lot! – pcalc Feb 2 '19 at 13:03
• @pcalc You are welcome. – saz Feb 2 '19 at 13:05