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I need some help on the following problem:

Let $X_1$ and $X_2$ be random sample from the pdf \begin{equation} f(x) = \begin{cases} 4x^3,&0<x<1\\ 0, & \text{otherwise} \end{cases} \end{equation} Obtain $P(X_1X_2\geq 1/4)$.

So here is what I did:

$P(X_1X_2\geq 1/4)=1-P(X_1X_2< 1/4)$

Next, I need to obtain the distribution of $X_1X_2$ in order to evaluate the above probability, but how? I really appreciate if you could give me a solution on this.

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You don't need to obtain the distribution of the product $X_1X_2$ to solve the problem.

The probability is determined by integrating the joint density function over the slice of the unit square above the hyperbola $xy=1/4$.

$$\begin{align} \mathsf P(X_1X_2 \geq \tfrac 1 4) & = \int_{1/4}^1 f(x)\;\mathsf P(X_2 \geq \tfrac1{4x}\mid X_1=x)\, \operatorname d x \\[1ex] & = \int_{1/4}^1 \int_{1/4x}^1 f(x)f(y)\,\operatorname d y\operatorname d x \\[1ex] & = 16 \int_{1/4}^1 \int_{1/4x}^1 x^3\; y^3\,\operatorname d y\operatorname d x \\[1ex] & = \end{align}$$

Can you take it from here?

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The distribution of $X_1X_2$ is $f(x,y)=16x^3y^3$, where I've used $x$ for $X_1$ and $y$ for $X_2$. The limits are just $0<x,y<1$ as before.

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  • $\begingroup$ Thanks, but what you did there is you obtain the joint distribution of ($X_1$, $X_2$), so to obtain the distribution of $X_1\cdot X_2$, I need to do bivariate transformation by letting $U = X_1\cdot X_2$ and $V = X_2$, after that I will compute the marginal distribution of $U$ and that would be the distribution of $X_1\cdot X_2$. I think that is how it should be done. $\endgroup$ – Al-Ahmadgaid Asaad Dec 13 '14 at 6:47
  • $\begingroup$ @Al-AhmadgaidAsaad If you want to do it that way, you can, but the distribution I gave can be used to solve the problem. I think the fault is in the question rather than the answer. $\endgroup$ – Suzu Hirose Dec 13 '14 at 7:55

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