Let $X$ and $Y$ be continuous random variables with probability density functions $P_X(x)$ and $P_Y(y)$, respectively. Further, let $Y = X^2$ and $P_X(x) = \begin{cases} 1, & x\in (0,1] \\ 0, & \text{otherwise} \end{cases}$
Which one of the following options is correct?
$P_Y(y) = \begin{cases} \frac{1}{2\sqrt{y}}, & y\in (0,1] \\ 0, & otherwise \end{cases} $
Solution: Finding the PDF of Y = X^2
We are given the probability density function (PDF) of a continuous random variable X:
\( P_X(x) = \begin{cases} 1, & x \in (0,1] \\ 0, & \text{otherwise} \end{cases} \)
We are also given the relationship \( Y = X^2 \). We need to find the PDF of Y, denoted as \( P_Y(y) \).
1. Determine the Range of Y
Since \( X \) is defined on \( (0, 1] \), and \( Y = X^2 \), the range for \( Y \) is found by squaring the bounds of \( X \):
\( 0^2 < Y \le 1^2 \)
Therefore, \( Y \) is defined on the interval \( (0, 1] \).
2. Calculate the Cumulative Distribution Function (CDF) of Y
The CDF of \( Y \) is \( F_Y(y) = P(Y \le y) \).
Substitute \( Y = X^2 \):
\( F_Y(y) = P(X^2 \le y) \)
Since \( X > 0 \) and we are considering \( y > 0 \), this is equivalent to:
\( F_Y(y) = P(X \le \sqrt{y}) \)
This equals the CDF of \( X \) evaluated at \( \sqrt{y} \), i.e., \( F_X(\sqrt{y}) \).
First, find the CDF of \( X \), \( F_X(x) \). For \( x \in (0, 1] \):
\( F_X(x) = \int_{-\infty}^{x} P_X(t) dt = \int_{0}^{x} 1 dt = [t]_0^x = x \)
So, \( F_X(x) = x \) for \( x \in (0, 1] \).
Now, substitute \( \sqrt{y} \) into \( F_X(x) \). Since \( y \in (0, 1] \), \( \sqrt{y} \) is also in \( (0, 1] \).
\( F_Y(y) = F_X(\sqrt{y}) = \sqrt{y} \quad \text{for } y \in (0, 1] \)
3. Derive the PDF of Y from its CDF
The PDF \( P_Y(y) \) is the derivative of the CDF \( F_Y(y) \):
\( P_Y(y) = \frac{d}{dy} F_Y(y) \)
For \( y \in (0, 1] \):
\( P_Y(y) = \frac{d}{dy} (\sqrt{y}) = \frac{d}{dy} (y^{1/2}) = \frac{1}{2} y^{(1/2 - 1)} = \frac{1}{2} y^{-1/2} = \frac{1}{2\sqrt{y}} \)
Therefore, the PDF of \( Y \) is:
\( P_Y(y) = \begin{cases} \frac{1}{2\sqrt{y}}, & y \in (0,1] \\ 0, & \text{otherwise} \end{cases} \)
Important Questions from Random Variables
Consider the following probability density function for a random variable x.
(a) \(f_{1}(x)=1\ ;\ -1\le x\le 1\)
(b) \(f_{2}(x)=x\ ;\ 0\le x\le 1\)
(c) \(f_{3}(x)=(1-x)\ ;\ -1\le x\le 1\)
Arrange the above functions in terms of the increasing value of mean of random variable x.
Considering all the symbols with their usual meanings, match the following :
List - I List - II (a) \(f_{X}(x)\) (i) \(\displaystyle\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dx\) (b) \(\displaystyle\int_{-\infty}^{\infty}f_{X}(x)\,dx\) (ii) \(\displaystyle\int_{-\infty}^{a}f_{X}(x)\,dx\) (c) \(F_{X}(a)\) (iii) \(\displaystyle\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dy\) (d) \(f_{Y}(y)\) (iv) 1 Codes :
The density function of a random variable is given by \(p(x)=Ke^{-\frac{x^{2}}{2}}\) for \(-\infty \lt x \lt \infty\). The value of K should be :
If the odds in favour of any random event A are 5 ∶ 6, then the odds against the event are:
If random variable X follows binomial distribution with parameter n and p with mean 15 and variance 10, then the value of mode is
