Task 1: Simulating a triangle

An isosceles right triangle has side length uniformly distributed on the unit interval (0, 1). Simulate 500 hypotenuses for such an isosceles right triangle.
Code set-up
runif(500,0,1) will generate 500 variates from a uniform(0, 1) distribution. Recall that for an isosceles right triangle, the hypotenuse is √2 times the length of one of the sides. You can use this formula to simulate the hypotenuse length from a simulated side legnth.
To plot the triangles, you need merely to plot the hypotenuse on a line using the lines function. If we stored our uniform random variates in a vector x, the following code chunk will plot the ith hypotenuse generated. This single hypotenuse is not very exciting. But if you wrap this code in a for-loop over the 500 simulation experiments, it will fill in the “triangle space”.
Note that this code has the eval=FALSE option just to present the code without output. Your code will not use this option.
# set up the plot by outlining the border from 0 to 1 on each of the triangle side 1 and side 2
plot(c(0,1,0),c(1,0,0), type="p", xlab="Triangle side 1", ylab="Triangle side 2")
i = 17 # plot the 17th hypotenuse; remove this line once you set-up the for-loop over i
# draw the hypotenuse
lines(c(0,x[i]),c(x[i],0))
The problem: Simulate 500 hypotenuses for the isosceles right triangle
# Dobrow Exercise 6.57, based on 6.8
# True expected value and variance are sqrt(2)/2 and 1/6

Report the following:
The empirical (simulated) expected length and variance of the length of the hypotenuse.
Since a side has length X∼U(0, 1), the hypotenuse H=X⋅√2 has expected value E(H)=√2⋅E(X)=√2/2. VAR(H)=2VAR(X)=1/6. Compare the truth with your empirical values in the previous bullet.
Present a histogram of the simulated hypotenuse lengths. What distribution does it seem to follow and why?
Present, on a single set of axes, the simulated hypotenuses using the triangle_plot code chunk above. That is, plot a diagonal line representing the hypotenuse for each simulated value. What do you find?

Task 2: Simulating Gaussian probabilities
Finding areas under the normal curve (integrating with respect to the normal pdf) is analytically quite difficult due to the form of the pdf. However, Monte Carlo estimates via simulation are quick and accurate.
The Box-Muller algorithm is a popular Gaussian simulator, applying a transformation method to convert uniform variates into normal variates. The algorithm for simulating two random variates Z_1 and Z_2 from a N(0, 1) distribution is as follow.

Box-Muller Gaussian simulator
Generate U_1,U_2∼U(0,1), two uniform random variates.
Compute Z_1=√(-2ln(U₁)) cos(2πU₂) and Z_2=√(-2ln(U₁)) sin(2πU₂).
Recall that if we want a variate X∼N(μ,σ^2), we can compute X=σZ+μ for both Z₁ and Z₂.

Code set-up
For those who want to try, the Box-Muller Gaussian simulator can be simulated without for-loops by using a matrix of generated uniform variates. Otherwise, code each of the items in the algorithm to generate two variates. Then wrap that in a for-loop to repeat and generate the desired number of Gaussian variates.
The following code will draw a histogram of your variates, x, along with a density plot for evaluation.
hist(x, xlab="N(2,2) variates", main="", freq=FALSE)
lines(seq(-4,8,0.05), dnorm(seq(-4,8,0.05),2,sqrt(2)), col="blue")
The problem
Use the Box-Muller algorithm to simulate 1000 random variates of X∼N(2, 2).

Report the following:
Mean, standard deviation, and median of the variates.
Histogram of the variates and an overlay with the true density.
The proportion of simulated values out of the 1000 that lie one, two, and three standard deviations from the mean (that is, within μ±σ, μ±2σ, and μ±3σ). These percentages are estimates of P(μ-iσ<X<μ+iσ) for i=1,2,3.
The true values of the three probabilities in (c) are 0.68, 0.95, and 0.997 respectively. Compare your estimates from (c) with the true values. This is called the ``68-95-99.7" rule for the Gaussian/Normal distribution.

Task 1: Simulating a triangle An isosceles right triangle has si...