During Class

Brain Gains

1. For the exponential distribution $f_3(x; \lambda) = \lambda e^{-\lambda x}$ with $\lambda >0 $ and $x\geq 0$ (and 0 otherwise), write down integrals that give the expected value and variance.

Solution

The general formulas are $\ds E[X] = \int_{-\infty}^\infty x f(x) dx$ and $\ds Var[X] = \int_{-\infty}^\infty (x-E[X]])^2 f(x) dx$. We update the bounds $0\leq x<\infty$ and function $f(x) = \lambda e^(-\lambda x)$ to obtain

$\ds E[X] = \int_0^\infty x (\lambda e^{-\lambda x}) dx$
$\ds Var[X] = \int_{0}^\infty (x-E[X]])^2 (\lambda e^{-\lambda x}) dx$.

2. Compute both the integrals above with Mathematica.

Solution

From the prep, we have the following code. The Assumptions command allows us to inform Mathematica about the fact that $\lambda>0$.

$Assumptions = \[Lambda] > 0;
f = \[Lambda]*Exp[-\[Lambda]*x];
bounds = {x, 0, Infinity};
A = Integrate[f, bounds]
EV = Integrate[x*f, bounds]
Var = Integrate[(x - EV)^2*f, bounds]
StDev = Sqrt[Var]
% // Simplify
$Assumptions = Null;

3. For the uniform distribution $f_0(x; a,b) = \frac{1}{b-a}$ for $a < x < b$ and 0 otherwise, write down integrals that give the expected value and variance, and then compute these integrals.

Solution

We update the bounds to $a\leq x\leq b$ and function $f(x) = \frac{1}{b-a}$ to obtain

$\ds E[X] = \int_a^b x (\frac{1}{b-a}) dx$
$\ds Var[X] = \int_a^b (x-E[X]])^2 (\frac{1}{b-a}) dx$.

The code chunk below computes these values, with some "%//Simplify" commands after the results to make them easier to work with later.

$Assumptions = a <= b;
f = 1/(b - a);
bounds = {x, a, b};
A = Integrate[f, bounds]
EV = Integrate[x*f, bounds]
% // Simplify
Var = Integrate[(x - EV)^2*f, bounds]
% // Simplify
StDev = Sqrt[Var]
% // Simplify
$Assumptions = Null;

Group Meeting

Solving a System of Equations

Solve the system $ \left\{\begin{array}{ll}\frac{1}{2}(x+y) = 2 \\ \frac{1}{12}(x-y)^2 = 3\end{array} \right.$ for $x$ and $y$, assuming $x<y$.
Solve the system $ \left\{\begin{array}{ll}\frac{1}{2}(a+b) = 2.7 \\ \frac{1}{12}(a-b)^2 = 10.2\end{array} \right.$ for $a$ and $b$, assuming $a<b$.

Have you seen the quantities $\frac{1}{2}(a+b)$ and $\frac{1}{12}(a-b)^2$ somewhere before?

Activity - More Practice with finding PDFs, Expected Value, Variance, and CDF

1. Let $g(x) = \begin{cases}e^{-4x} & x \geq 0 \\ 0 & \text{otherwise}\end{cases}$.

Find $k$ so that $f(x) = k g(x)$ is a PDF for a random variable $X$.
Write down the definite integrals that give the expected value and variance of $X$. Then use Mathematica to compute the expected value and variance of $X$.
Write down the definite integral that gives $F(x)$, the cumulative distribution function for $X$. Then use Mathematica to compute $F(x)$ for $x\geq 0$.
Compute $P(X\leq 2)$ and then $P(X\geq 2)$.
Find a value $c$ so that $P(X\leq c)=0.90$ (we call this the 90th percentile).
By hand, compute $F'(x)$ and compare it to $f(x)$.

Solution

We get $A = \int_{-\infty}^{\infty}g(x)dx = \int_{0}^{\infty}e^{-4x}dx = \frac{1}{4}$, which means $k=\frac{1}{A} = 4$ and $f(x) = \begin{cases}4e^{-4x} & 0\leq x<\infty\\ 0 & \text{otherwise}\end{cases}$.
We have $E[X] = \int_{-\infty}^{\infty}xf(x)dx = \int_{0}^{\infty}x4e^{-4x}dx = \frac{1}{4}$ and $\text{Var}[X] = \int_{-\infty}^{\infty}(x-E[X])^2f(x)dx = \int_{0}^{\infty}(x-\frac{1}{4})^2 4e^{-4x}dx = \frac{1}{16}$.
We have $F(x) = \int_{-\infty}^{x}f(x)dx$. If $x<0$ then $F(x)=0$, but for $x\geq 0$ we have $F(x) = \int_{-\infty}^{x}f(x)dx =\int_{0}^{x}4e^{-4x}dx = 1-e^{-4x}$.
The values are $P(X\leq 10) = F(10) = 1-e^{-8}$ and then $P(X\geq 10)=1-F(10) = e^{-8}$.
We need to solve $0.90 = 1-e^{-4x}$. This means $e^{-4x} = 0.1$ which gives $-4x = \ln(0.1)$ or $x = \ln(0.1)/(-4)$.
The derivative is $F'(x) = 0-e^{-4x}(-4) = 4e^{-4x}=f(x)$ for $x\geq 0$ (and zero otherwise).

The Mathematica code computes all of the above. Feel free to adapt this code as you tackle other problems.

g = Exp[-4*x];
a = 0;
b = Infinity;
bounds = {x, a, b};
A = Integrate[g, bounds]
k = 1/A
f = k g
EV = Integrate[x f, bounds]
Var = Integrate[(x - EV)^2 f, bounds]
F = Integrate[f, {x, a, x}]
Integrate[f, {x, a, 2}]
1-Integrate[f, {x, a, 2}]
Solve[Integrate[f, {x, a, c}]==0.9, c]
D[F, x](*Computes derivative of F with respect to x. Should match f*)

2. Let $g(x) = \begin{cases}e^{-\lambda x} & x \geq 0\\ 0 & \text{otherwise}\end{cases}$ with $\lambda >0$. In Mathematica, the following code tells the computer that $\lambda$ is positive, which will simplifies the output quite a bit.

g = Exp[-\[Lambda] x]
$Assumptions = \[Lambda] > 0

Find $k$ so that $f(x) = k g(x)$ is a PDF for a random variable $X$. We call this an exponential random variable.
Write down the definite integrals that give the expected value and variance of $X$. Then use Mathematica to compute the expected value and variance of $X$.
Write down the definite integral that gives $F(x)$, the cumulative distribution function for $X$. Then use Mathematica to compute $F(x)$ for $x\geq 0$.
By hand, compute $F'(x)$ and compare it to $f(x)$.

3. Let $g(x) = \begin{cases}1 & -3\leq x\leq 5\\ 0 & \text{otherwise}\end{cases}$.

Find $k$ so that $f(x) = k g(x)$ is a PDF for a random variable $X$.
Write down the definite integrals that give the expected value and variance of $X$. Then use Mathematica to compute the expected value and variance of $X$.
Write down the definite integral that gives $F(x)$, the cumulative distribution function for $X$. Then use Mathematica to compute $F(x)$ for $-3\leq x\leq 5$. For $x<-3$, what is $F(x)$? For $x>5$, what is $F(x)$?
By hand, compute $F'(x)$ and compare it to $f(x)$.

4. Let $g(x) = \begin{cases}1 & a\leq x\leq b\\ 0 & \text{otherwise}\end{cases}$ where $a<x<b$.

Find $k$ so that $f(x) = k g(x)$ is a PDF for a random variable $X$. We call this a uniform random variable.
Write down the definite integrals that give the expected value and variance of $X$. Then use Mathematica to compute the expected value and variance of $X$.
Write down the definite integral that gives $F(x)$, the cumulative distribution function for $X$. Then use Mathematica to compute $F(x)$ for $a\leq x\leq b$. For $x<a$, what is $F(x)$? For $x>b$, what is $F(x)$?
By hand, compute $F'(x)$ and compare it to $f(x)$.

5. Let $g(x) = \begin{cases}3 & -2 \leq x < 1\\ 5 & 1 \leq x \leq 5\\ 0 & \text{otherwise}\end{cases}$.

Find $k$ so that $f(x) = k g(x)$ is a PDF for a random variable $X$.
Write down the definite integrals that give the expected value and variance of $X$. Then use Mathematica to compute the expected value and variance of $X$.
Compute $F(x)$.
By hand, compute $F'(x)$ and compare it to $f(x)$.

Solution

We can use piecewise function notation in Mathematica. Here is an example. We need the assumptions command that $x$ is a real number to compute the CDF $F$ and actually have Mathematica do the integral.

$Assumptions = x \[Element] Reals
g = Piecewise[{{3, -2 <= x && x < 1}, {5, 1 <= x <= 5}, {0, True}}]
Plot[g, {x, -5, 10}]
A = Integrate[g, {x, -Infinity, Infinity}]
k = 1/A
f = k g
1 == Integrate[ f, {x, -Infinity, Infinity}]
EV = Integrate[x f, {x, -Infinity, Infinity}]
Var = Integrate[(x - EV)^2 f, {x, -Infinity, Infinity}]
F = Integrate[ f, {x, -Infinity, x}]
D[F, x]

Discussion

Expected Value and Variance of Common Distributions

We found the expected value and variance for 4 common probability distributions. Let's recap what we found.

For an exponential distribution, we have $E[X] = ...$ and $\text{Var}[X] = ...$. (Where did we compute these?)
For a uniform distribution, we have $E[X] = ...$ and $\text{Var}[X] = ...$. (Where did we compute these?)
For a normal distribution with mean $\mu$ and standard deviation $\sigma$, we have $E[X] = \mu$ and $\text{Var}[X] = \sigma^2$.
For a gamma distribution with shape $\alpha$ and rate $\beta$, we have $E[X] = \frac{\alpha}{\beta}$ and $\text{Var}[X] = \frac{\alpha}{\beta^2}$.

Method of Moments

The expected value and variance are identifying characteristics of a random variable. We can match these characteristics from the distribution with these characteristics of the data, and use this to determine unknown parameters in a model. Here are the steps:

Calculate the mean (expected value) of the distribution as a function of the parameters of the distribution.
Calculate the variance of the distribution as a function of the parameters of the distribution.
Set the mean of the distribution equal to the mean of the data.
Set the variance of the distribution equal to the variance of the data.
Solve the system of equations for the parameter values.

Note:

If the distribution (or model) has only one parameter, then skip steps 2 and 4 and solve the equation you find in step 3.
If the distribution (or model) has more than two parameters, calculate additional identifying characteristics of the distribution and data and set them equal and then solve the resulting system of equations.

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HomePage Schedule	Class 40 During Class Brain Gains 1. For the exponential distribution $f_3(x; \lambda) = \lambda e^{-\lambda x}$ with $\lambda >0 $ and $x\geq 0$ (and 0 otherwise), write down integrals that give the expected value and variance. Solution The general formulas are $\ds E[X] = \int_{-\infty}^\infty x f(x) dx$ and $\ds Var[X] = \int_{-\infty}^\infty (x-E[X]])^2 f(x) dx$. We update the bounds $0\leq x<\infty$ and function $f(x) = \lambda e^(-\lambda x)$ to obtain $\ds E[X] = \int_0^\infty x (\lambda e^{-\lambda x}) dx$ $\ds Var[X] = \int_{0}^\infty (x-E[X]])^2 (\lambda e^{-\lambda x}) dx$. 2. Compute both the integrals above with Mathematica. Solution From the prep, we have the following code. The Assumptions command allows us to inform Mathematica about the fact that $\lambda>0$. $Assumptions = \[Lambda] > 0; f = \[Lambda]Exp[-\[Lambda]x]; bounds = {x, 0, Infinity}; A = Integrate[f, bounds] EV = Integrate[xf, bounds] Var = Integrate[(x - EV)^2f, bounds] StDev = Sqrt[Var] % // Simplify $Assumptions = Null; 3. For the uniform distribution $f_0(x; a,b) = \frac{1}{b-a}$ for $a < x < b$ and 0 otherwise, write down integrals that give the expected value and variance, and then compute these integrals. Solution We update the bounds to $a\leq x\leq b$ and function $f(x) = \frac{1}{b-a}$ to obtain $\ds E[X] = \int_a^b x (\frac{1}{b-a}) dx$ $\ds Var[X] = \int_a^b (x-E[X]])^2 (\frac{1}{b-a}) dx$. The code chunk below computes these values, with some "%//Simplify" commands after the results to make them easier to work with later. $Assumptions = a <= b; f = 1/(b - a); bounds = {x, a, b}; A = Integrate[f, bounds] EV = Integrate[xf, bounds] % // Simplify Var = Integrate[(x - EV)^2f, bounds] % // Simplify StDev = Sqrt[Var] % // Simplify $Assumptions = Null; Group Meeting Solving a System of Equations Solve the system $ \left\{\begin{array}{ll}\frac{1}{2}(x+y) = 2 \\ \frac{1}{12}(x-y)^2 = 3\end{array} \right.$ for $x$ and $y$, assuming $x<y$. Solve the system $ \left\{\begin{array}{ll}\frac{1}{2}(a+b) = 2.7 \\ \frac{1}{12}(a-b)^2 = 10.2\end{array} \right.$ for $a$ and $b$, assuming $a<b$. Have you seen the quantities $\frac{1}{2}(a+b)$ and $\frac{1}{12}(a-b)^2$ somewhere before? Activity - More Practice with finding PDFs, Expected Value, Variance, and CDF 1. Let $g(x) = \begin{cases}e^{-4x} & x \geq 0 \\ 0 & \text{otherwise}\end{cases}$. Find $k$ so that $f(x) = k g(x)$ is a PDF for a random variable $X$. Write down the definite integrals that give the expected value and variance of $X$. Then use Mathematica to compute the expected value and variance of $X$. Write down the definite integral that gives $F(x)$, the cumulative distribution function for $X$. Then use Mathematica to compute $F(x)$ for $x\geq 0$. Compute $P(X\leq 2)$ and then $P(X\geq 2)$. Find a value $c$ so that $P(X\leq c)=0.90$ (we call this the 90th percentile). By hand, compute $F'(x)$ and compare it to $f(x)$. Solution We get $A = \int_{-\infty}^{\infty}g(x)dx = \int_{0}^{\infty}e^{-4x}dx = \frac{1}{4}$, which means $k=\frac{1}{A} = 4$ and $f(x) = \begin{cases}4e^{-4x} & 0\leq x<\infty\\ 0 & \text{otherwise}\end{cases}$. We have $E[X] = \int_{-\infty}^{\infty}xf(x)dx = \int_{0}^{\infty}x4e^{-4x}dx = \frac{1}{4}$ and $\text{Var}[X] = \int_{-\infty}^{\infty}(x-E[X])^2f(x)dx = \int_{0}^{\infty}(x-\frac{1}{4})^2 4e^{-4x}dx = \frac{1}{16}$. We have $F(x) = \int_{-\infty}^{x}f(x)dx$. If $x<0$ then $F(x)=0$, but for $x\geq 0$ we have $F(x) = \int_{-\infty}^{x}f(x)dx =\int_{0}^{x}4e^{-4x}dx = 1-e^{-4x}$. The values are $P(X\leq 10) = F(10) = 1-e^{-8}$ and then $P(X\geq 10)=1-F(10) = e^{-8}$. We need to solve $0.90 = 1-e^{-4x}$. This means $e^{-4x} = 0.1$ which gives $-4x = \ln(0.1)$ or $x = \ln(0.1)/(-4)$. The derivative is $F'(x) = 0-e^{-4x}(-4) = 4e^{-4x}=f(x)$ for $x\geq 0$ (and zero otherwise). The Mathematica code computes all of the above. Feel free to adapt this code as you tackle other problems. g = Exp[-4x]; a = 0; b = Infinity; bounds = {x, a, b}; A = Integrate[g, bounds] k = 1/A f = k g EV = Integrate[x f, bounds] Var = Integrate[(x - EV)^2 f, bounds] F = Integrate[f, {x, a, x}] Integrate[f, {x, a, 2}] 1-Integrate[f, {x, a, 2}] Solve[Integrate[f, {x, a, c}]==0.9, c] D[F, x](Computes derivative of F with respect to x. Should match f) 2. Let $g(x) = \begin{cases}e^{-\lambda x} & x \geq 0\\ 0 & \text{otherwise}\end{cases}$ with $\lambda >0$. In Mathematica, the following code tells the computer that $\lambda$ is positive, which will simplifies the output quite a bit. g = Exp[-\[Lambda] x] $Assumptions = \[Lambda] > 0 Find $k$ so that $f(x) = k g(x)$ is a PDF for a random variable $X$. We call this an exponential random variable. Write down the definite integrals that give the expected value and variance of $X$. Then use Mathematica to compute the expected value and variance of $X$. Write down the definite integral that gives $F(x)$, the cumulative distribution function for $X$. Then use Mathematica to compute $F(x)$ for $x\geq 0$. By hand, compute $F'(x)$ and compare it to $f(x)$. 3. Let $g(x) = \begin{cases}1 & -3\leq x\leq 5\\ 0 & \text{otherwise}\end{cases}$. Find $k$ so that $f(x) = k g(x)$ is a PDF for a random variable $X$. Write down the definite integrals that give the expected value and variance of $X$. Then use Mathematica to compute the expected value and variance of $X$. Write down the definite integral that gives $F(x)$, the cumulative distribution function for $X$. Then use Mathematica to compute $F(x)$ for $-3\leq x\leq 5$. For $x<-3$, what is $F(x)$? For $x>5$, what is $F(x)$? By hand, compute $F'(x)$ and compare it to $f(x)$. 4. Let $g(x) = \begin{cases}1 & a\leq x\leq b\\ 0 & \text{otherwise}\end{cases}$ where $a<x<b$. Find $k$ so that $f(x) = k g(x)$ is a PDF for a random variable $X$. We call this a uniform random variable. Write down the definite integrals that give the expected value and variance of $X$. Then use Mathematica to compute the expected value and variance of $X$. Write down the definite integral that gives $F(x)$, the cumulative distribution function for $X$. Then use Mathematica to compute $F(x)$ for $a\leq x\leq b$. For $x<a$, what is $F(x)$? For $x>b$, what is $F(x)$? By hand, compute $F'(x)$ and compare it to $f(x)$. 5. Let $g(x) = \begin{cases}3 & -2 \leq x < 1\\ 5 & 1 \leq x \leq 5\\ 0 & \text{otherwise}\end{cases}$. Find $k$ so that $f(x) = k g(x)$ is a PDF for a random variable $X$. Write down the definite integrals that give the expected value and variance of $X$. Then use Mathematica to compute the expected value and variance of $X$. Compute $F(x)$. By hand, compute $F'(x)$ and compare it to $f(x)$. Solution We can use piecewise function notation in Mathematica. Here is an example. We need the assumptions command that $x$ is a real number to compute the CDF $F$ and actually have Mathematica do the integral. $Assumptions = x \[Element] Reals g = Piecewise[{{3, -2 <= x && x < 1}, {5, 1 <= x <= 5}, {0, True}}] Plot[g, {x, -5, 10}] A = Integrate[g, {x, -Infinity, Infinity}] k = 1/A f = k g 1 == Integrate[ f, {x, -Infinity, Infinity}] EV = Integrate[x f, {x, -Infinity, Infinity}] Var = Integrate[(x - EV)^2 f, {x, -Infinity, Infinity}] F = Integrate[ f, {x, -Infinity, x}] D[F, x] Discussion Expected Value and Variance of Common Distributions We found the expected value and variance for 4 common probability distributions. Let's recap what we found. For an exponential distribution, we have $E[X] = ...$ and $\text{Var}[X] = ...$. (Where did we compute these?) For a uniform distribution, we have $E[X] = ...$ and $\text{Var}[X] = ...$. (Where did we compute these?) For a normal distribution with mean $\mu$ and standard deviation $\sigma$, we have $E[X] = \mu$ and $\text{Var}[X] = \sigma^2$. For a gamma distribution with shape $\alpha$ and rate $\beta$, we have $E[X] = \frac{\alpha}{\beta}$ and $\text{Var}[X] = \frac{\alpha}{\beta^2}$. Method of Moments The expected value and variance are identifying characteristics of a random variable. We can match these characteristics from the distribution with these characteristics of the data, and use this to determine unknown parameters in a model. Here are the steps: Calculate the mean (expected value) of the distribution as a function of the parameters of the distribution. Calculate the variance of the distribution as a function of the parameters of the distribution. Set the mean of the distribution equal to the mean of the data. Set the variance of the distribution equal to the variance of the data. Solve the system of equations for the parameter values. Note: If the distribution (or model) has only one parameter, then skip steps 2 and 4 and solve the equation you find in step 3. If the distribution (or model) has more than two parameters, calculate additional identifying characteristics of the distribution and data and set them equal and then solve the resulting system of equations. Edit Page History Source Attach File Backlinks List Group Page last modified on March 23, 2026, at 12:44 PM*
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40

During Class

Brain Gains

Group Meeting

Solving a System of Equations

Activity - More Practice with finding PDFs, Expected Value, Variance, and CDF

Discussion

Expected Value and Variance of Common Distributions

Method of Moments