Question

If \(U_{1}, U_{2}, U_{3}\) are independent uniform \((0,1)\) random variables, find \(P\left(\prod_{i=1}^{3} U_{i}>0.1\right)\) Hint: Relate the desired probability to one about a Poisson process.

Short answer

The desired probability is \(P\left(\prod_{i=1}^{3} U_{i}>0.1\right) = \boxed{\frac{1}{2}}\).

Step-by-step solution

Take the natural logarithm of the inequality

We need to find the probability \(P(\prod_{i=1}^{3} U_{i}>0.1)\). First, let's take the natural logarithms of both sides of the inequality \[\ln(\prod_{i=1}^{3} U_{i}) > \ln(0.1)\] Since \(\ln(xy) = \ln(x) + \ln(y)\) for random variables \(x\) and \(y\), we get: \[\ln(U_1) + \ln(U_2) + \ln(U_3) > \ln(0.1)\]

Define Poisson process for each logarithm

We will now define a new set of variables \(Y_{1}, Y_{2}, Y_{3}\) as: \[Y_i = -\ln(U_i)\] We are given that \(U_{1}\), \(U_{2}\), and \(U_{3}\) are independent, so the \(Y_i\) are also independent. Now, our inequality becomes: \[-Y_1 - Y_2 - Y_3 > -\ln(0.1)\] Taking the negation of this inequality: \[Y_1 + Y_2 + Y_3 < \ln(10)\] Now we need to evaluate the cumulative distribution functions \(F_{Y_i}(y)\) for \(i = 1, 2, 3\).

Find the cumulative distribution functions of new random variables

To find the cumulative distribution functions of \(Y_i\), we will first find their probability density functions: For \(y \geq 0\), \[F_{Y_i}(y) = P(Y_i \leq y) = P(-\ln(U_i) \leq y) \] \[= P(\ln(U_i) \geq -y) = P(U_i \geq e^{-y})\] Since \(U_i\) is a uniform random variable, \[F_{U_i}(u) = u\] Now let's substitute this into the above equation: \[F_{Y_i}(y) = P(U_i \geq e^{-y}) = 1 - F_{U_i}(e^{-y}) = 1 - e^{-y}\] To find the probability density function \(f_{Y_i}(y)\), we can differentiate the cumulative distribution function: \[f_{Y_i}(y) = \frac{d}{dy} F_{Y_i}(y) = e^{-y}\]

Find the joint distribution function

Since \(Y_1\), \(Y_2\), and \(Y_3\) are independent, we can find the joint distribution function by multiplying their individual distributions: \[f_{Y_1,Y_2,Y_3}(y_1,y_2,y_3) = f_{Y_1}(y_1)f_{Y_2}(y_2)f_{Y_3}(y_3) = e^{-y_1}e^{-y_2}e^{-y_3} = e^{-(y_1+y_2+y_3)}\] Now we can find the desired probability: \[P(Y_1+Y_2+Y_3 < \ln(10)) = \int_0^{\ln(10)} \int_0^{\ln(10)-y_1} \int_0^{\ln(10)-y_1-y_2} f_{Y_1,Y_2,Y_3}(y_1,y_2,y_3) dy_3 dy_2 dy_1\]

Evaluate the integral

Now, let's evaluate the integral to find the desired probability: \[P(Y_1+Y_2+Y_3 < \ln(10)) = \int_0^{\ln(10)} \int_0^{\ln(10)-y_1} \int_0^{\ln(10)-y_1-y_2} e^{-(y_1+y_2+y_3)} dy_3 dy_2 dy_1\] \[=\int_0^{\ln(10)} \int_0^{\ln(10)-y_1} e^{-(y_1+y_2)}\left[ 1 - e^{-(\ln(10)-y_1-y_2)} \right] dy_2 dy_1\] \[=\int_0^{\ln(10)} \int_0^{\ln(10)-y_1} \left[e^{-(y_1+y_2)} - e^{-\ln(10)}\right] dy_2 dy_1\] \[= 1 - e^{-\ln(10)}\int_0^{\ln(10)} \int_0^{\ln(10)-y_1} dy_2 dy_1\] \[= 1 - e^{-\ln(10)}\int_0^{\ln(10)} (\ln(10)-y_1) dy_1 \] \[= 1 - e^{-\ln(10)}\left[\ln^2(10)-\frac{1}{2}\ln^2(10)\right] = 1 - \frac{1}{2} = \frac{1}{2}\] Therefore, the desired probability is \(P\left(\prod_{i=1}^{3} U_{i}>0.1\right) = \boxed{\frac{1}{2}}\).