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: Density of $Y=X^2$ (A).
With $x=\sqrt y$, $f_Y(y)=f_X(\sqrt y)\frac1{2\sqrt y}$. On $0<y<1$ this is $\frac1{4\sqrt y}$, on $1<y<9$ it is $\frac1{8\sqrt y}$. (A) holds.

Step 2: Density of $Z=1/X$ (B).
With $x=1/z$, $f_Z(z)=f_X(1/z)\frac1{z^2}$. For $\frac13<z<1$ it is $\frac1{4z^2}$, for $z>1$ it is $\frac1{2z^2}$. (B) holds.

Step 3: Finiteness of means (C).
$Y$ is bounded in $(0,9)$, so $E(Y)<\infty$. But $E(Z)=E(1/X)$ includes $\int_0^1\frac1x\cdot\frac12 dx$, which diverges. So $E(Z)$ is infinite and (C) fails.

Step 4: Skewness of $Y$ (D).
$Y=X^2$ has a long right tail out to $9$, so it is positively skewed. (D) holds.

Step 5: Collect.
\[ \boxed{(A),(B),(D)} \]