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 $c$.

(b) Find the probability $P(X> 2c)$.

Solution.

Solution of (a)

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*}

The post Condition that a Function Be a Probability Density Function first appeared on Problems in Mathematics.

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 of $T$ with respect to the basis $B$. For $f(x)\in P_2(\R)$, define $T$ as follows.

(a) \[T(f(x))=\frac{\mathrm{d}^2}{\mathrm{d}x^2} f(x)-3\frac{\mathrm{d}}{\mathrm{d}x}f(x)\]

(b) \[T(f(x))=\int_{-1}^1\! (t-x)^2f(t) \,\mathrm{d}t\]

(c) \[T(f(x))=e^x \frac{\mathrm{d}}{\mathrm{d}x}(e^{-x}f(x))\]

Hint.

To find the matrix, we just need to compute $T(1), T(x), T(x^2)$ and read the coefficients of $1, x, x^2$.

Solution.

(a) $T(f(x))=\frac{\mathrm{d}^2}{\mathrm{d}x^2} f(x)-3\frac{\mathrm{d}}{\mathrm{d}x}f(x)$

Since the derivative the constant function $1$ is zero, we have
\[T(1)=0.\] Since the second derivative of the function $x$ is zero, we have
\[T(x)=-3.\] Finally, we calculate
\[T(x^2)=2-6x.\] There with respect to the basis $\{1, x, x^2\}$, the matrix for $T$ is
\[\begin{bmatrix}
0 & -3 & 2 \\
0 &0 &-6 \\
0 & 0 & 0
\end{bmatrix}.\]

(b) $T(f(x))=\int_{-1}^1\! (t-x)^2f(t) \,\mathrm{d}t$

We first compute
\begin{align*}
T(f(x))&=\int_{-1}^{1} \! (t^2-2tx-+x^2)f(t) \, \mathrm{d}t \\
&=\left( \int_{-1}^{1} \! t^2f(t) \, \mathrm{d}t\right)-2\left( \int_{-1}^{1} \! tf(t) \, \mathrm{d}t \right) x+\left(\int_{-1}^{1} f(t) \, \mathrm{d}t \right) x^2.
\end{align*}
Then we have
\begin{align*}
T(1)&=\frac{2}{3} +2x^2 \\
T(x)&=-\frac{4}{x} x\\
T(x^2)&=\frac{2}{5}+\frac{2}{3}x^2
\end{align*}

Here we calculated the integrals
\[\int_{-1}^{1} \! 1 \, \mathrm{d}t= \left[t \right]_{-1}^1=2, \,\,\,\, \int_{-1}^{1} \!t \, \mathrm{d}t=\left[\frac{1}{2}t^2\right]_{-1}^1=0,\] \[\int_{-1}^{1} \! t^2 \, \mathrm{d}t=\left[\frac{1}{3}t^3 \right]_{-1}^1=\frac{2}{3}, \,\,\,\, \int_{-1}^{1} \! t^3 \, \mathrm{d}t=\left[\frac{1}{4}t^4 \right]_{-1}^1=0, \,\,\,\, \int_{-1}^{1} \! t^4 \, \mathrm{d}t=\left[\frac{1}{5}t^5 \right]_{-1}^1=\frac{2}{5}.
\] Therefore, the matrix for the linear transformation with respect to the basis $B=\{1,x,x^2\}$ is
\[\begin{bmatrix}
\frac{2}{3} & 0 & \frac{2}{5} \\[6pt] 0 &-\frac{4}{3} &0 \\[6pt] 2 & 0 & \frac{2}{3}
\end{bmatrix}.\]

(c) $T(f(x))=e^x \frac{\mathrm{d}}{\mathrm{d}x}(e^{-x}f(x))$

 Note that by the product rule for derivatives,
\begin{align*}
T(f(x))&=e^x \left(-e^{-x}f(x)+e^{-x}\frac{\mathrm{d}}{\mathrm{d}x} f(x) \right)\\
&=-f(x)+\frac{\mathrm{d}}{\mathrm{d}x} f(x).
\end{align*}
Thus we have
\begin{align*}
T(1)&=-1\\
T(x)&=-x+1\\
T(x^2)&=-x^2+2x.
\end{align*}
Hence the matrix for the linear transformation $T$ with respect to the basis $B$ is
\[\begin{bmatrix}
-1 & 1 & 0 \\
0 &-1 &2 \\
0 & 0 & -1
\end{bmatrix}.\]

Comment.

This problem hints a connection between linear algebra and calculus, or differential equations.
And yes, they are closely related and differential equations can be solved using techniques of linear algebras.

The post Matrix Representations for Linear Transformations of the Vector Space of Polynomials first appeared on Problems in Mathematics.