Condition that a Function Be a Probability Density Function
Problem 756
Let $c$ be a positive real number. Suppose that $X$ is a continuous random variable whose probability density function is given by
\begin{align*}
f(x) = \begin{cases}
\frac{1}{x^3} & \text{ if } x \geq c\\
0 & \text{ if } x < c.
\end{cases}
\end{align*}
(a) Determine the value of $c$.
As $f(x)$ is a probability density function of a continuous random variable $X$, its integral must sum to 1, that is,
\[\int_{-\infty}^{\infty} f(x) dx = 1.\]
(This follows from $1=P(X\in (-\infty, \infty)) = \int_{-\infty}^{\infty} f(x) dx$.)
As $f(x)=1/x^3$ when $x \geq c$ and $f(x) = 0$ when $x < c$, we have
\begin{align*}
\int_{-\infty}^{\infty} f(x) dx & = \int_{-\infty}^c f(x) dx + \int_{c}^{\infty} f(x) dx\\[6pt]
&=0 + \int_{c}^{\infty} \frac{1}{x^3} dx\\[6pt]
&=\left[\frac{x^{-2}}{-2}\right]_c^{\infty}\\[6pt]
&= \frac{1}{2c^2}.
\end{align*}
Since this must be equal to $1$, we obtain
\[1 = \frac{1}{2c^2}\]
and thus
\[c = \frac{1}{\sqrt{2}},\]
as $c$ is positive.
Solution of (b)
In part (a), we found $c = 1/ \sqrt{2}$, hence $2c=\sqrt{2}$.
As $f(x)$ is the probability density function of the continuous random variable $X$, we have
\[P(X > \sqrt{2}) = \int_{\sqrt{2}}^\infty f(x) dx.\]
So, the desired probability can be computed as follows.
\begin{align*}
P(X>\sqrt{2}) &= \int_{\sqrt{2}}^{\infty} \frac{1}{x^3} dx\\[6pt]
&= \left[\frac{x^{-2}}{-2}\right]_{\sqrt{2}}^{\infty}\\[6pt]
&= \frac{1}{4}.
\end{align*}
Expectation, Variance, and Standard Deviation of Bernoulli Random Variables
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 $X$ […]
Linearity of Expectations E(X+Y) = E(X) + E(Y)
Let $X, Y$ be discrete random variables. Prove the linearity of expectations described as
\[E(X+Y) = E(X) + E(Y).\]
Solution.
The joint probability mass function of the discrete random variables $X$ and $Y$ is defined by
\[p(x, y) = P(X=x, Y=y).\]
Note that the […]
How to Prove Markov’s Inequality and Chebyshev’s Inequality
(a) Let $X$ be a random variable that takes only non-negative values. Prove that for any $a > 0$,
\[P(X \geq a) \leq \frac{E[X]}{a}.\]
This inequality is called Markov's inequality.
(b) Let $X$ be a random variable with finite mean $\mu$ and variance $\sigma^2$. Prove that […]
How to Use the Z-table to Compute Probabilities of Non-Standard Normal Distributions
Let $X\sim \mathcal{N}(\mu, \sigma)$ be a normal random variable with parameter $\mu=6$ and $\sigma^2=4$. Find the following probabilities using the Z-table below.
(a) Find $P(X \lt 7)$.
(b) Find $P(X \lt 3)$.
(c) Find $P(4.5 \lt X \lt 8.5)$.
The Z-table is […]
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Let $X$ and $Y$ be geometric random variables with parameter $p$, with $0 \leq p \leq 1$. Assume that $X$ and $Y$ are independent.
Let $n$ be an integer greater than $1$. Let $k$ be a natural number with $k\leq n$. Then prove the formula
\[P(X=k \mid X + Y = n) = […]
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Let $P_2(\R)$ be the vector space over $\R$ consisting of all polynomials with real coefficients of degree $2$ or less.
Let $B=\{1,x,x^2\}$ be a basis of the vector space $P_2(\R)$.
For each linear transformation $T:P_2(\R) \to P_2(\R)$ defined below, find the matrix representation […]
Upper Bound of the Variance When a Random Variable is Bounded
Let $c$ be a fixed positive number. Let $X$ be a random variable that takes values only between $0$ and $c$. This implies the probability $P(0 \leq X \leq c) = 1$. Then prove the next inequality about the variance $V(X)$.
\[V(X) \leq \frac{c^2}{4}.\]
Proof.
Recall that […]
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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) […]