Let $X$ be an exponential random variable with parameter $\lambda$.

(a) For any positive integer $n$, prove that
\[E[X^n] = \frac{n}{\lambda} E[X^{n-1}].\]

(b) Find the expected value of $X$.

(c) Find the variance of $X$.

(d) Find the standard deviation of $X$.

Solution.

Solution of (a)

Recall that the probability density function $f(x)$ of an exponential random variable with parameter $\lambda$ is given by
\begin{align*}
f(x) =
\begin{cases}
\lambda e^{-\lambda x} & \text{ if } x \geq 0\\
0 & \text{ if } x < 0 \end{cases} \end{align*} and the parameter $\lambda$ must be positive. It follows from this and the definition of expectation, we get \begin{align*} E[X^n] &= \int_0^{\infty} x^n \cdot \lambda e^{-\lambda x} dx. \end{align*} Applying integral by parts with \[u = x^n, dv=\lambda e^{-\lambda x} dx\] and hence \[du = nx^{n-1}dx, v = -e^{-\lambda x},\] we obtain (from $\int u dv = uv - \int v du$) \begin{align*} E[X^n] &= \int_0^{\infty} x^n \cdot \lambda e^{-\lambda x} dx\\[6pt] &= \left[x^n \cdot (-e^{-\lambda x})\right]_0^{\infty} - \int_0^{\infty} (-e^{\lambda x}) \cdot nx^{n-1} dx\\[6pt] &= 0 + n \int_0^{\infty} e^{\lambda x} x^{n-1} dx\\[6pt] &= \frac{n}{\lambda} \int_0^{\infty} x^{n-1} \cdot \lambda e^{\lambda x} dx\\[6pt] &= \frac{n}{\lambda} E[X^{n-1}]. \end{align*} This proves the required equality \[E[X^n] = \frac{n}{\lambda} E[X^{n-1}].\]

Solution of (b)

The expected value $E[X]$ can be obtained from the formula we just proved in part (a) by substituting $n=1$. Thus, we have
\begin{align*}
E[X] = \frac{1}{\lambda} E[1] = \frac{1}{\lambda}.
\end{align*}

Solution of (c)

We calculate the variance using the formula
\[V(X) = E[X^2] – (E[X])^2.\] We know $E[X] = 1/\lambda$ from part (b). To compute $E[X^2]$, let $n=2$ in the formula in part (a). Then
\begin{align*}
E[X^2] &= \frac{2}{\lambda}E[X]\\[6pt] &= \frac{2}{\lambda} \cdot \frac{1}{\lambda}\\[6pt] &= \frac{2}{\lambda^2}.
\end{align*}
Combining these, we obtain
\begin{align*}
V(X) &= E[X^2] – (E[X])^2\\[6pt] &= \frac{2}{\lambda^2} – \frac{1}{\lambda^2}\\[6pt] & = \frac{1}{\lambda^2}.
\end{align*}
Therefore, the variance of $X$ is
\[V(X) = \frac{1}{\lambda^2}.\]

Solution of (d)

Taking the square root of the variance $V(X)$, we obtain the standard deviation
\[\sigma = \frac{1}{\lambda}.\] Click here if solved 4

The post Expected Value and Variance of Exponential Random Variable first appeared on Problems in Mathematics.

A box of some snacks includes one of five toys. The chances of getting any of the toys are equally likely and independent of the previous results.

(a) Suppose that you buy the box until you complete all the five toys. Find the expected number of boxes that you need to buy.

(b) Find the variance and the standard deviation of the event in part (a).

Solution.

Solution of (a)

Let $X$ be the number of boxes that you need to buy until you complete all the five toys. Our goal is to compute the expected value $E[X]$.
To achieve this, we consider the next random variables. Let $X_i$ be the number of boxes you need to buy to get $i$th toy after getting $i-1$ toys. Then it is clear from definition that
\[X = X_1 + X_2 + X_3 + X_4 + X_5.\] For example, $X_1$ is the number of boxes you need to buy to get the first toy. Since whenever you open the first box, it is guaranteed that you get a new toy, we have $X_1 = 1$.
Also, to get the second toy after the first one, there are $4/5$ chance of getting new toy and $1/5$ chance of getting the same toy as the first one. Thus, $X_2$ is a geometric random variable with parameter $4/5$. We denote this as $X \sim G_{4/5}$. Similarly, we get
\[X_3 \sim G_{3/5}, \quad X_4 \sim G_{2/5}, \quad \text{ and } X_5 \sim G_{1/5}.\]

By the linearity of expectation, we have
\begin{align*}
E[X] &= E[X_1 + X_2 + X_3 + X_4 + X_5]\\
&= E[X_1] + E[X_2] + E[X_3] + E[X_4] + E[X_5]\\
&= E[1] + E[G_{4/5}] + E[G_{3/5}] + E[G_{2/5}] + E[G_{1/5}] \end{align*}
Now, the expected value of a geometric random variable $G_p$ is given by
\[E[G_p] = \frac{1}{p}.\] It follows that
\begin{align*}
E[X] &= 1 + \frac{5}{4} +\frac{5}{3} + \frac{5}{2} + \frac{5}{1}\\
&= 5\left(1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}\right)\\
&\approx 11.41667
\end{align*}
Thus, the expected number of boxes that you need to buy to complete all the five toys is 11.41667.

Solution of (b)

Now we compute the variance of $X$. Recall that the variance of a geometric random variable $G_p$ is given by
\[V(G_p) = \frac{1-p}{p^2}.\] As we have seen that
\begin{align*}
X &= X_1 + X_2 + X_3 + X_4 + X_5\\
&\sim 1 + G_{4/5} + G_{3/5} + G_{2/5} + G_{1/5}
\end{align*}
and as each random variable $X_i$ is independent of each other, we obtain
\begin{align*}
V(X) &= V(1 + G_{4/5} + G_{3/5} + G_{2/5} + G_{1/5})\\[6pt] &= V(1) + V(G_{4/5}) + V(G_{3/5}) + V(G_{2/5}) + V(G_{1/5})\\[6pt] &= 0 + \frac{1-\frac{4}{5}}{\left(\frac{4}{5}\right)^2} + \frac{1-\frac{3}{5}}{\left(\frac{3}{5}\right)^2} + \frac{1-\frac{2}{5}}{\left(\frac{2}{5}\right)^2} + \frac{1-\frac{1}{5}}{\left(\frac{1}{5}\right)^2}\\[6pt] &\approx 25.17361
\end{align*}
Thus, the variance is $V(X) = 25.17361$.

The standard deviation is the square root of the variance. Hence, we obtain
\[\sigma(X) \approx \sqrt{25.17361} \approx 5.01733.\]

Remark.

This type of problems is called a Coupon Collecting Problem.

The post Coupon Collecting Problem: Find the Expectation of Boxes to Collect All Toys first appeared on Problems in Mathematics.

A final exam of the course Probability 101 consists of 10 multiple-choice questions. Each question has 4 possible answers and only one of them is a correct answer. To pass the course, 8 or more correct answers are necessary. Assume that a student has not studied probability at all and has no idea how to solve the questions. So the student decided to answer each questions randomly. Thus, for each of 10 questions, the student choose one of the 4 answers randomly and each choice is independent each other.

(1) What is the probability that the student answered correctly only one question among the 10 questions?

(2) Determine the probability that the student passes the course.

(3) What is the expected value of the number of questions the student answered correctly?

(4) Find the variance and standard deviation of the number of questions the student answered correctly.

Solution.

Solution of (1)

Let $X$ be the number of questions that the student answered correctly. For each question, the probability that the student select a correct answer is $1/4$. As selections are independent, the random variable $X$ is binomial with parameters $n=10$ and $p=1/4$. Let $P(i)$ denote the probability that the student answered $i$ questions correctly.
Then the probability that the student selects only one correct answer is given by
\[P(0) = {10 \choose 1} \left(\frac{1}{4}\right)^1 \left(\frac{3}{4}\right)^9 \approx 0.1877\]

Solution of (2)

To pass the course, 8 or more correct answers are must. Thus,
\begin{align*}
&P(\text{6 or more correct answers}) = P(8) + P(9) + P(10)\\[6pt] &= {10 \choose 8} \left(\frac{1}{4}\right)^8 \left(\frac{3}{4}\right)^2 +
{10 \choose 9} \left(\frac{1}{4}\right)^9 \left(\frac{3}{4}\right)^1 +
{10 \choose 10} \left(\frac{1}{4}\right)^{10} \left(\frac{3}{4}\right)^0 \\
&\approx 0.0003862380 + 0.0000286102 + 0.0000009536 \\
&= 0.0004158018 = 0.04158018 \%.
\end{align*}

Hence, the desired probability is 0.0004158018 or 0.04158018%.

Solution of (3)

Recall that the expected value of a binomial random variable $X$ with parameters $(n, p)$ is given by
\[E[X] = np.\] As $n=10$ and $p=1/4$, the expected value of the number of questions the student answered correctly is
\[E[X] = 10 \cdot \frac{1}{4} = 2.5.\]

Solution of (4)

Recall that the variance and the standard deviation of a binomial random variable $X$ with parameters $(n, p)$ are given by
\[V(X) = np(1-p) \text{ and } \sigma(X) = \sqrt{np(1-p)}.\] Substituting $n=10$ and $p=1/4$, we obtain the variance
\begin{align}
V(X) &= 10 \cdot \frac{1}{4} \cdot \frac{3}{4} = \frac{15}{8} = 1.875
\end{align}
and the standard deviation
\begin{align*}
\sigma(X) &= \sqrt{1.875} \approx 1.3693.
\end{align*}

The post Can a Student Pass By Randomly Answering Multiple Choice Questions? first appeared on Problems in Mathematics.

A random variable $X$ is said to be a Bernoulli random variable if its probability mass function is given by
\begin{align*}
P(X=0) &= 1-p\\
P(X=1) & = p
\end{align*}
for some real number $0 \leq p \leq 1$.

(1) Find the expectation of the Bernoulli random variable $X$ with probability $p$.

(2) Find the variance of $X$.

(3) Find the standard deviation of $X$.

Solution.

Solution of (1)

As $X$ is a Bernoulli random variable, it takes only two values $0$ or $1$.
Thus, by definition of expectation, we obtain
\begin{align*}
E[X] &= \sum_{i=0}^1 P(X=i)x\\
&= P(X=0) \cdot 0 + P(X=1) \cdot 1\\
&= (1-p) \cdot 0 + p \cdot 1\\
&= p.
\end{align*}
Hence, the expectation of the Bernoulli random variable $X$ with parameter $p$ is $E[X] = p$.

Solution of (2)

We calculate the variance of the Bernoulli random variable $X$ using the definition of a variance. Namely, the variance of $X$ is defined as
\[V(X) = E[X^2] – \left(E[X]\right)^2.\] Here is an observation that makes the computation simpler: As the Bernoulli random variable takes only the values $0$ or $1$, it follows that $X^2 = X$. Thus, the variance can be computed as follows.
\begin{align*}
V(X) &= E[X^2] – \left(E[X]\right)^2 && \text{by definition of variance}\\
&= E[X] – \left(E[X]\right)^2 && \text{by observation $X^2=X$}\\
&= p – p^2 && \text{by result of (1)}\\
&= p(1-p)
\end{align*}

Thus, the variance of the Bernoulli random variable $X$ with parameter $p$ is given by
\[V(X) = p(1-p).\]

Solution of (3)

The standard deviation is obtained by taking the square root of the variance. Hence, using the result of (2), the standard deviation of the Bernoulli random variable $X$ with parameter $p$ is
\[\sigma(X) = \sqrt{p(1-p)}.\]

Related Problem

For a solution, see the post Given the Variance of a Bernoulli Random Variable, Find Its Expectation

The post Expectation, Variance, and Standard Deviation of Bernoulli Random Variables first appeared on Problems in Mathematics.