Class 37

Between Class Sessions - Prep for Day 37

Please spend around 2 hours working between class sessions, focusing on the tasks below. Use any extra time to complete KnewtonAlta assignments and/or work on Project tasks.

Pick something from your prep today that you can share with your group in class. It might be something new that you learned. It might be questions you have that are still unanswered, or a question along with what helped you eventually answer it. It might be something tricky that you solved. It might be a review topic that helped you remember something. It might be a conversation you had with AI that was helpful. You will have a chance to share this with your peers during class. Come ready to articulate your thinking and questions.

Preparation

(1) Recap Reading - Rectangles, Targets, and Sums

For Unit 3, the reading as well as many practice exercises appear in the document Rectangles, Targets, and Sums

Skim read the entire document above (it’s a second pass over the material). Make notes of questions you have that you would like to discuss more. If you didn’t watch the video in section 1.2 the first time through, please do so this time through.
Complete a few exercises that perhaps you skipped before, or have a better understanding of now. Come ready to share what you did with your group. Remember it’s OK if you don’t have the correct answers. The point is to have made a good faith attempt, and then come ready to share what you did.
Please focus on the following exercises.
- 1.9 #4
- 2.1 #4
- 2.2 #4
- 2.3 #4
- 3.2 #1
- 4.2 #2 (pick one or two)
- 4.3 #1 (pick one)
One of the main goals of doing the exercises above is to focus on why \(P(a < X \leq b) = P(a \leq X \leq b) = F(b)-F(a)\). Write a sentence or two that explains why this is true, based off your work from the exercises above.

Regular Reminders

Skill Practice (KA Homework)

Continue working on 3 – Riemann Sums & Definite Integrals

Applied Practice (Project Work)

Finish and Submit Project 3 Task 1

During Class

Brain Gains

draw_target Code

draw_target <- function(f,a,b,num_points=100){
  x <- c(a,seq(a,b,(b-a)/num_points),b,a)
  y <- c(0,f(seq(a,b,(b-a)/num_points)),0,0)
  par(mar=c(2.5,2.5,0.25,0.25))
  plot(x,y,type = "l")
  polygon(x,y,col="gray")
}

Use the two properties of probability density functions to determine if \(f\) a PDF, where \[f(x) = \begin{cases} 5-x & \quad 2 \leq x \leq 5 \\ \\ 0 & \text{otherwise}. \end{cases}\]

f <- function(x){5-x}
draw_target(f,2,5)

Consider the target with top defined by \(f\). When a dart hits the point \((x , y )\), we’ll record the \(x\)-coordinate and let \(X\) represent this random variable. A PDF for this random variable is \[f(x) = \begin{cases} kx - k & \quad 1 \leq x \leq 3 \\ \\ 0 & \quad \text{otherwise}. \end{cases}\] Compute the value of \(k\).

f <- function(x){x-1}
draw_target(f,1,3)

Discussion

Riemann Sums with R, Definite integrals with Mathematica

draw_rect_approx() code

draw_rect_approx <- function(f,a,b,num_rectangles, method = "mid"){
  n <- num_rectangles
  dx <- (b-a)/n
  x <- c(a,seq(a,b,dx/100),b,a)
  y <- c(0,f(seq(a,b,dx/100)),0,0)
  par(mar=c(2.5,2.5,0.25,0.25))
  plot(x,y,type = "l")

  if(method == "left"){
    xi <- seq(a+0*dx/2,b-dx/2,dx)
    lines(xi,f(xi),type = "h")
    lines(xi,f(xi),type = "s")
    lines(c(xi[n],xi[n]+dx),f(c(xi[n],xi[n])),type = "l")
    lines(c(xi[n],xi[n]+dx),f(c(xi[n],xi[n])),type = "h")
  }
  else if(method == "right"){
    xi <- seq(a+dx,b+dx/2,dx)
    lines(xi-dx,f(xi),type = "h")
    lines(xi-dx,f(xi),type = "s")
    lines(c(xi[n]-dx,xi[n]),f(c(xi[n],xi[n])),type = "l")
    lines(c(xi[n]-dx,xi[n]),f(c(xi[n],xi[n])),type = "h")
  } 
  else{#Use midpoint
    xi <- seq(a+dx/2,b,dx)
    lines(xi-dx/2,f(xi),type = "h")
    lines(xi-dx/2,f(xi),type = "s")
    lines(c(xi[n]-dx/2,xi[n]+dx/2),f(c(xi[n],xi[n])),type = "l")
    lines(c(xi[n]-dx/2,xi[n]+dx/2),f(c(xi[n],xi[n])),type = "h")
  }
}

Let’s recap Riemann Sums and Definite Integrals.

Let \(g(x) = x^2e^{-x}\) for \(1\leq x\leq 4\), and zero otherwise. The code below draws the function \(g\) and then uses 6 rectangles to approximate the area under \(g\) where the height of each rectangle is the value of the function at the right endpoint of the rectangle.

g <- function(x){x^2*exp(-x)}
a <- 1
b <- 4
n <- 6
draw_rect_approx(g,a,b,n,method = "right")
dx <- (b-a)/n
xi <- seq(a+dx,b,dx)  # Can you tell what part of this line of code uses right endpoints? a+dx
data.frame(right_point = xi, function_at_xi = g(xi), area_i = g(xi)*dx)
c(riemann_sum_using_right_endpoints = sum(g(xi)*dx))

We can increase the number of rectangles for a better approximation. Here we use 10 million rectangles in the Riemann \(\sum g(x_i)\Delta x\) to approximate the area under \(g\).

n <- 10000000 # Don't plot anymore
dx <- (b-a)/n
xi <- seq(a+dx,b,dx)  
A <- sum(g(xi)*dx)
A

Because the area under \(g\) is not 1, we multiply by \(k=\frac{1}{A}\) to obtain the probability density function \(f(x) = k g(x)\) of a random variable \(X\). Let’s check the area under \(f\) is 1 using a Riemann sum \(\sum f(x_i)\Delta x\).

k <- 1/A
f <- function(x){k*g(x)}
sum(f(xi)*dx) #Should be 1.

We now approximate the expected value and variance of \(X\) using

\(E[X]\approx \sum x_if(x_i)\Delta x\) and
\(\text{Var}[X]\approx\sum (x_i-E[X])^2f(x_i)\Delta x\). Variance is a measure of how spread out a distribution is.

EV <- sum(xi*f(xi)*dx)
EV
Var <- sum((xi-EV)^2*f(xi)*dx)
Var

Because the area under \(f\) is 1, when we compute probabilities we don’t have to divide by area. Let’s compute \(P(X\leq 3)\) using 10 million rectangles by letting \(b=3\).

a <- 1
b <- 3 # Use 3 instead of 4.
n <- 10000000
dx <- (b-a)/n
xi <- seq(a+dx,b,dx)  
sum(f(xi)*dx) # Probability won't be 1 anymore.

With the PDF \(f\), we compute \(P(2\leq X\leq 3)\) by updating the lower and upper bounds.

a <- 2 # Use 2 instead of 1
b <- 3 # Use 3 instead of 4
n <- 10000000
dx <- (b-a)/n
xi <- seq(a+dx,b,dx)  
sum(f(xi)*dx)

We now define the function \(P(a,b) = P(a\leq X\leq b)\) and \(F(x) = P(X\leq x)\) to simplify the computations above.

P <- function(a,b,n=10000000){
  dx <- (b-a)/n
  xi <- seq(a+dx,b,dx)  
  sum(f(xi)*dx)
}
F <- function(x,a=1){P(a,x)}
P(1,4)  
P(1,3)  
F(3)

Why must we have \(P(2\leq X\leq 3) = F(3)-F(2)\)?

P(2,3)
F(3)-F(2)
all.equal(P(2,3),F(3)-F(2))

Now let’s swap to Mathematica and use definite integral notation \(\int_a^b f(x) dx\) to do the same computations. Remember, a definite integral is just the symbol for letting the number of rectangles approach infinity, written symbolically as \[\int_a^b f(x) dx =\lim_{n\to \infty} \sum_{i=1}^nf(x_i)\Delta x_i\]

We first find the area under \(g\). (Remember to use shift + enter to run the command in Mathematica.)

g = x^2 Exp[-x];
a = 1
b = 4
A = Integrate[g, {x, a, b}]
N[A, 7] (*This provides 7 significant figures*)
Plot[g, {x, a, b}]

With the area under \(g\), we obtain a probability density function \(f = kg\) and check that the area under f is actually 1.

k = 1/A;
f = k*g;
Integrate[f, {x, a, b}]

We now compute the expected value and variance of a random variable with pdf of \(f(x)\). We’ll give both exact and approximate answers to compare with R.

EV = Integrate[x*f, {x, a, b}]
EV // N (*This is another example of code to obtain a numerical approximation. Typing N[EV] gives the same result.*)
Var = Integrate[(x - EV)^2*f, {x, a, b}]
Var // N

We can quickly compute various probabilities by adjusting the bounds of the definite integral. The probabilities \(P(X\leq 3)\) and \(P(2\leq X\leq 3)\) are given below.

Integrate[f, {x, a, 3}]
% // N
Integrate[f, {x, 2, 3}]
% // N

Mathematica has no problems working with symbols, something R is not designed to do. We can compute the cumulative distribution function \(F(x)\) using Mathematica, as well as verify that \(F'(x) = f(x)\).

F = Integrate[f, {x, a, x}] (*Make the right bound x, rather than a number*)
D[F, x] (*Computes F'(x)*)
f (*Compare to the previous*)
f - D[F, x] // FullSimplify (*See if the difference is zero*)

Group Meeting

Practice with some other functions

Work together on your computers to help one another complete the following problems.

For each function below, use Mathematica to compute

the area \(A = \int_a^b g(x) dx\),
the value \(k\) so that \(f=kg\) is a PDF for a random variable,
the expected value \(E[X] =\int_a^b xf(x)dx\), and
the variance \(\text{Var}[X]=\int_a^b (x-E[X])^2f(x)dx\).

Let \(g(x) = \sqrt{49-x^2}\) for \(0\leq x\leq 7\) - use “g=Sqrt[49-x^2]”. The region under \(g(x)\) is a quarter of the circle \(x^2+y^2=49\). You will see \(\pi\) show up in lots of answers in Mathematica. Use “%//N” to approximate the output.
Let \(g(x) = \frac{\ln(x+1)}{x^2}\) for \(1\leq x\leq 5\) - use “g=Log[x+1]/x^2”.
Let \(g(x) = e^{-3x}\) for \(0\leq x\leq \infty\) - use “g=Exp[-3x]”. You can enter Infinity as a value for the upper bound.

Some Solutions

\(A\approx 38.4845\), \(E[X]\approx 2.97089\), \(\text{Var}[X]\approx 3.4238\).
\(A\approx 0.845621\), \(E[X]\approx 2.27858\), \(\text{Var}[X]\approx 1.15167\).
\(A=\frac{1}{3}\), \(E[X]=\frac{1}{3}\), \(\text{Var}[X]=\frac{1}{9}\).

Computing Probabilities using a Probability Density function

Work on the chalkboard as you complete these problems, and pass the chalk as you finish each one. Leave up your work so that you can compare your answers with your neighbors. If you notice an answer differs from a neighboring group’s, then have a discussion with them.

Consider the function \(g(x) = 2x\) for \(0\leq x\leq 5\), and 0 otherwise.

Find the area under \(g\) using a geometric argument. Then compute \(A=\int_0^5 2x dx\) with Mathematica to verify you have the same answer. Then let \(k=\frac{1}{A}\) so that \(f(x) = k g(x)\) is a probability density function for a random variable \(X\).
Explain why \(P(X\leq 2) = \frac{4}{25}\) by using a geometric argument. Then use Mathematica to verify that \(\int_0^2 f(x) dx = \frac{4}{25}\).
Explain why \(P(X\leq 3) = \frac{9}{25}\) by using a geometric argument. Then use Mathematica to verify that \(\int_0^3 f(x) dx = \frac{9}{25}\).
Explain why \(P(2<X\leq 3) = P(X\leq 3) - P(X\leq 2)\).
Explain why \(P(2<X\leq 3) = P(2\leq X\leq 3)\).
Explain why \(P(2\leq X\leq 3) = \int_2^3 f(x) dx\).
Use Mathematica to show that \(P(1.2\leq X\leq 3.7) = 0.49\).
Use Mathematica to show that \(P(X\leq x) = \frac{x^2}{25}\).
Use Mathematica to show that \(P(a\leq X\leq b) = \frac{b^2}{25}-\frac{a^2}{25}\), provided \(0\leq a\leq b\leq 5\).
Explain why \(\int_a^b f(x) dx = F(b)-F(a)\).

If you finish early, swap to the function \(g(x) = 7-x\) for \(0\leq x\leq 4\) and 0 otherwise.

Find the area under \(g\) using a geometric argument. Then compute \(A=\int_0^4 g dx\) with Mathematica to verify you have the same answer. Then let \(k=\frac{1}{A}\) so that \(f(x) = k g(x)\) is a probability density function for a random variable \(X\).
Compute \(P(X\leq 2)\) using a geometric argument. Then use Mathematica to verify your result.
Compute \(P(X\leq 3)\) using a geometric argument. Then use Mathematica to verify your result.
Compute \(P(2\leq X\leq 3)\) using a geometric argument. Then use Mathematica to verify your result.
Verify that \(\int_2^3 f(x) dx = P(X\leq 3) - P(X\leq 2)\) by computing each side with Mathematica.
Use Mathematica to compute \(F(x) = P(X\leq x)\). [Hint, compute \(\int_0^x f(x) dx\).]
Use Mathematica to compute \(P(a\leq X\leq b)\). [Hint, compute \(\int_a^b f(x) dx\).]
Explain why \(\int_a^b f(x) dx = F(b)-F(a)\).

Discussion

We’ll discuss some of the above, if/as needed.

Source: Class.37 on byuimath.com