Can somebody give me hint to solve the following problem: Let $Y1,Y2,Y3, \ldots$ be a sequence of positive i.i.d. random variables with $0 < E[\ln Y_i]=\gamma<\infty$. Define the sequence ${X_n,n=1,2,3,...}$ as\begin{equation} X_n=(Y_1 Y_2 Y_3 \cdots Y_{n-1}Y_n)^{\frac{1}{n}}, \qquad \textrm{ for }n=1,2,3,\cdots. \end{equation}show that $X_n \xrightarrow{p} e^{\gamma}$.
My attempt: By continuous mapping theorem, $X_n \xrightarrow{p} e^{\gamma} \implies Z_n = \ln X_n \xrightarrow{p} \gamma$. Now,
$P(|Z_n-\gamma|) \geq \epsilon ) \leq \frac{Var(Z_n)}{\epsilon^2}$, by Chebyshev's ineqality. The question is: How to bound the variance?
The weak law of large numbers implies $\frac{1}{n} \sum_{i=1}^n \ln Y_i \overset{p}{\to} \gamma$. The continuous mapping theorem then implies $X_n = e^{\frac{1}{n} \sum_{i=1}^n \ln Y_i} \overset{p}{\to} e^\gamma$. The weak law of large numbers does not require that the variance of $\ln Y_i$ is finite.