Question

The Hit-Miss Method: Suppose \(g\) is bounded in \([0,1]-\) for instance, suppose \(0 \leqslant g(x) \leqslant b\) for \(x \in[0,1]\). Let \(U_{1}, U_{2}\) be independent random numbers and set \(X=U_{1}, Y=b U_{2}\) -so the point \((X, Y)\) is uniformly distributed in a rectangle of length 1 and height \(b\). Now set $$ I=\left\\{\begin{array}{ll} 1, & \text { if } Y

Short answer

(a) We have calculated the expectation using the joint probability density function of independent random variables \(X\) and \(Y\), which are uniformly distributed: \(E[bI] = \int_0^1 \int_0^{g(x)} b \cdot f_X(x) \cdot f_Y(y) dy dx = \int_0^1 \int_0^{g(x)} (b \cdot 1 \cdot \frac{1}{b}) dy dx = \int_0^1 g(x) dx\). (b) By calculating the variances of \(bI\) and \(g(U)\) and comparing them, we have shown: \(\operatorname{Var}(b I) \geq \operatorname{Var}(g(U))\), which indicates that the hit-miss method has a larger variance than simply computing \(g\) of a random number.

Step-by-step solution

(a) Calculate expected value (Expectation) of bI

To calculate \(E[b I]\), we will use the definition of expectation for a continuous random variable: \(E[bI] = \int_{-\infty}^{\infty} bI \cdot f_{(X,Y)}(x, y) d(x, y)\) where \(f_{(X,Y)}(x, y)\) represents the joint probability density function (PDF) of the random variables \(X\) and \(Y\). Since \(X\) and \(Y\) are independent random variables, we know that: \(f_{(X,Y)}(x, y) = f_X(x) \cdot f_Y(y)\) Both \(X\) and \(Y\) are uniformly distributed. Thus, their PDFs are: \(f_X(x) = \left\\{\begin{array}{ll}1, & 0 \leq x \leq 1 \\0, & \text{otherwise}\end{array}\right.\) \(f_Y(y) = \left\\{\begin{array}{ll}\frac{1}{b}, & 0 \leq y \leq b \\0, & \text{otherwise}\end{array}\right.\) Now, following the definition of \(I\), we compute \(E[bI]\): \(E[bI] = \int_0^1 \int_0^{g(x)} b \cdot f_X(x) \cdot f_Y(y) dy dx = \int_0^1 \int_0^{g(x)} (b \cdot 1 \cdot \frac{1}{b}) dy dx = \int_0^1 g(x) dx\). Therefore, we have shown \(E[b I]=\int_{0}^{1} g(x) dx\).

(b) Compare Var(bI) and Var(g(U))

To compare the variances, we first need to calculate \(\operatorname{Var}(b I)\) and \(\operatorname{Var}(g(U))\). For \(bI\): \(\operatorname{Var}(b I) = E[b^2 I^2] - (E[b I])^2\) To find \(E[b^2 I^2]\), we can use a similar approach as in part (a): \(E[b^2 I^2] = \int_0^1 \int_0^{g(x)} (b^2 \cdot 1 \cdot \frac{1}{b}) dy dx = b \int_0^1 g(x) dx = b \cdot E[b I] \). Now, we substitute \(E[b^2 I^2]\) back into the formula for \(\operatorname{Var}(b I)\): \(\operatorname{Var}(b I) = b \cdot E[b I] - (E[b I])^2\). For \(g(U)\): \(\operatorname{Var}(g(U)) = E[g(U)^2] - (E[g(U)])^2\). We know from part (a) that \(E[g(U)] = E[b I]\). Now we need to find \(E[g(U)^2]\): \(E[g(U)^2] = \int_0^1 g^2(x) dx\) Since \(g^2(x) \geq 0 \) for \(0 \leq x \leq 1\), we have \(E[g(U)^2] \geq (E[g(U)])^2\). Now, we substitute this back into the formula for \(\operatorname{Var}(g(U))\): \(\operatorname{Var}(g(U)) = E[g(U)^2] - (E[g(U)])^2 \leq E[g(U)^2]\) Considering the \(\operatorname{Var}(b I)\) and \(\operatorname{Var}(g(U))\) expressions found earlier, we have \(\operatorname{Var}(b I) = b \cdot E[bI] - (E[bI])^2 \geq E[g(U)^2] \geq \operatorname{Var}(g(U))\). Thus, we have shown that \(\operatorname{Var}(b I) \geq \operatorname{Var}(g(U))\), which means the hit-miss method has a larger variance than simply computing \(g\) of a random number.

Key concepts

Random Variables

In the realm of statistics and probability, random variables are a pivotal concept that allow mathematicians to quantify randomness. A random variable is essentially a variable whose possible values represent outcomes of a random phenomenon. For instance, when you roll a six-sided dice, the outcome is a random variable that can take any value between 1 and 6, each with an equal chance of occurring.

Random variables are classified into two categories: discrete and continuous. Discrete random variables have a countable number of possible values. Think of them like individual steps on a staircase. On the other hand, continuous random variables have an infinite number of potential values, analogous to a ramp, where you can stand at any point.

In the context of the Hit-Miss Method, the example utilizes continuous random variables represented by points within a rectangle on the coordinate plane. The location of any point picked at random within this rectangle can be described by the random variables X and Y. This approach can be used in computational methods to determine areas or to simulate random events governed by a certain probability distribution.

Probability Density Function

Moving deeper into the world of probabilities, the Probability Density Function (PDF) is a function that describes the likelihood of a random variable taking on a particular value. It's like a map that shows how the probabilities are distributed across the range of possible values. For continuous random variables, the PDF provides the probabilities in a continuous manner — hence the name density.

For any given continuous random variable, the probability that it falls within a certain range is given by the area under the curve of the PDF over that range. The Hit-Miss method uses PDFs to evaluate probabilities over a geometrical space. By integrating the PDF across a specific range, we obtain the probability for an event within that range.

The uniform distribution PDFs were used in the solution for both X and Y because each point within the given rectangle is equally likely to occur — every location is just as good as any other, with no bias towards any particular point.

Expected Value

The Expected Value is the average outcome you would anticipate after many trials of a random process. If you think of it in terms of a game, it's what you'd predict as your average earnings after playing the game many times. It is a crucial concept because it provides a measure of central tendency for random variables, much like the mean does for a sample of data.

In the first part of the exercise, we calculated the expected value of the indicator function bI, which was set to one or zero depending on whether the point (X, Y) fell under the curve g(X). The expected value was found by integrating over the region where Y is less than g(X), essentially quantifying the probability-weighted area under the curve. This agreed with the area calculation, which is intuitively what you'd expect if you repeatedly chose points at random and checked if they were below the curve.

Variance

The notion of Variance is fundamental in understanding the spread of a distribution. While expected value tells us where the center of our distribution lies, variance tells us how spread out the values are around that center. It's a measure of variability: low variance means that the data points tend to be close to the expected value, while high variance indicates they are more spread out.

Part (b) of the exercise compared the variance of the hit-miss method to that of simply computing the evaluation of the function g at a random number. The comparison revealed that the hit-miss method has a higher variance, which in practice means that it may be less stable or require more samples to achieve the same level of accuracy or precision as direct computation.

Variance is especially important in simulations and sampling processes because it impacts the reliability of the estimates. A method with lower variance will typically provide more consistent estimates than one with higher variance if the number of trials or samples is the same.