Question

Let \(X\) be a random variable with the probability density function

• A. A probability density function of \(Y\) is given by \[ f_Y(y)= \begin{cases} \dfrac{1}{4\sqrt{y}}, & 0<y<1,\\ \dfrac{1}{8\sqrt{y}}, & 1<y<9,\\ 0, & \text{otherwise} \end{cases} \]
• B. A probability density function of \(Z\) is given by \[ f_Z(z)= \begin{cases} \dfrac{1}{4z^2}, & \dfrac{1}{3}<z<1,\\ \dfrac{1}{2z^2}, & 1<z<\infty,\\ 0, & \text{otherwise} \end{cases} \]
• C. \(E(Y)<\infty\) and \(E(Z)<\infty\)
• D. The distribution of \(Y\) is positively skewed

Hint:

For transformation of variables, use \(f_Y(y)=f_X(x)\left|\frac{dx}{dy}\right|\), after expressing \(x\) in terms of the new variable.

Answer 1

The Correct Option is A, B, D

Step 1: Find the density of \(Y=X^2\).
Since \(X>0\), we have
\[ Y=X^2 \] So,
\[ x=\sqrt{y} \] and
\[ \left|\frac{dx}{dy}\right|=\frac{1}{2\sqrt{y}} \] Therefore,
\[ f_Y(y)=f_X(\sqrt{y})\frac{1}{2\sqrt{y}} \] For \(0<y<1\), we have \(0<\sqrt{y}<1\), so
\[ f_X(\sqrt{y})=\frac12 \] Thus,
\[ f_Y(y)=\frac12\cdot \frac{1}{2\sqrt{y}} =\frac{1}{4\sqrt{y}} \] For \(1<y<9\), we have \(1<\sqrt{y}<3\), so
\[ f_X(\sqrt{y})=\frac14 \] Thus,
\[ f_Y(y)=\frac14\cdot \frac{1}{2\sqrt{y}} =\frac{1}{8\sqrt{y}} \] Hence, option (A) is true.

Step 2: Find the density of \(Z=\frac{1}{X}\).
Let
\[ Z=\frac{1}{X} \] Then,
\[ x=\frac{1}{z} \] and
\[ \left|\frac{dx}{dz}\right|=\frac{1}{z^2} \] Since \(0<X<3\), we have
\[ Z>\frac13 \] Now, if
\[ 1<X<3, \] then
\[ \frac13<Z<1 \] and
\[ f_X\left(\frac1z\right)=\frac14 \] Therefore,
\[ f_Z(z)=\frac{1}{4z^2},\qquad \frac13<z<1 \] If
\[ 0<X<1, \] then
\[ Z>1 \] and
\[ f_X\left(\frac1z\right)=\frac12 \] Therefore,
\[ f_Z(z)=\frac{1}{2z^2},\qquad 1<z<\infty \] Hence, option (B) is true.

Step 3: Check finiteness of \(E(Y)\) and \(E(Z)\).
Since
\[ Y=X^2 \] and \(X\) is bounded between \(0\) and \(3\), we have
\[ 0<Y<9 \] Hence,
\[ E(Y)<\infty \] Now,
\[ E(Z)=E\left(\frac1X\right) \] \[ E(Z)=\int_0^1 \frac1x\cdot \frac12\,dx+\int_1^3 \frac1x\cdot \frac14\,dx \] But
\[ \int_0^1 \frac1x\,dx=\infty \] Therefore,
\[ E(Z)=\infty \] Hence, option (C) is false.

Step 4: Check skewness of \(Y\).
For \(Y=X^2\), the distribution has a long right tail up to \(9\).
Also, the third central moment is positive.
Therefore, the distribution of \(Y\) is positively skewed.
Hence, option (D) is true.

Step 5: Final conclusion.
The true statements are
\[ \boxed{(A),\,(B)\text{ and }(D)} \]