Joint density function of $f_\theta(x) = \theta x^{\theta-1}$

probability distributions

Let $f_\theta (x) = \theta x^{\theta-1}$, where $0<x<1$, $x_1, x_2$ are iid
Find $P(x_1x_2 \geq 0.75)$

I used double integration to obtain the joint probablity,
$$\begin{align*}
P(x_1x_2 \geq 0.75) &= \int_{0.75}^1 \theta x_1^{\theta – 1} \int_{0.75/x_1}^1 \theta x_2^{\theta – 1} dx_2 dx_1 \\
&= \dots \text{skip the steps of integration} \\
&= 1 – 0.75^\theta + 0.75^\theta \cdot ln0.75
\end{align*}$$

However, refer to For the density function $f_\theta (x) = \theta x^{\theta-1}$, how can we find the product distribution, $\prod_{i=1}^n X_i$?
Should $P(x_1x_2 \geq 0.75) = P(Gamma(2,\theta) \geq 0.75)$?
The answers are different using two approaches.

Update
I think I got it,
$$\begin{align*}
P(x_1x_2 \geq 0.75) &= P(exp(-Gamma(2,\theta)) \geq 0.75) \\
&= P(Gamma(2,\theta) \leq -ln0.75)
\end{align*}$$

Is this right?

Best Answer

to get a gamma distribution you transformed $X_1X_2$ in a certain way...you have to do the same with the requested quantile...

$$P(G< -\log0.75)$$

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