Finding Probability Density Function (PDF) from Distribution Function
For a continuous random variable $X$, the Probability Density Function (PDF), often denoted as $p(x)$ or $f_X(x)$, is mathematically derived from its Cumulative Distribution Function (CDF), denoted here as $F(x)$ (or $f(x)$ in the question). The relationship is fundamental in probability theory.
CDF to PDF Relationship
The Cumulative Distribution Function (CDF) $F(x)$ gives the probability that the random variable $X$ takes on a value less than or equal to $x$, i.e., $F(x) = P(X \le x)$.
The Probability Density Function (PDF) $p(x)$ describes the relative likelihood for this random variable to take on a given value. For a continuous random variable, the PDF is the derivative of the CDF.
Therefore, if $f(x)$ represents the CDF:
$ p(x) = \frac{d}{dx} f(x) $Analyzing the Options
- Option 1: $\frac{d}{dx} f(x)$ - This represents the derivative of the distribution function $f(x)$. This matches the definition of the PDF derived from the CDF.
- Option 2: $\frac{d^2}{dx^2} f(x)$ - This is the second derivative of the CDF, not the PDF.
- Option 3: $\int f(x)dx$ - This is the indefinite integral of the function $f(x)$. If $f(x)$ were the PDF, its integral would give the CDF.
- Option 4: $\int \frac{1}{f(x)} dx$ - This integral does not represent the PDF or the CDF.
Conclusion
The correct option describing the Probability Density Function (PDF) is the derivative of the distribution function $f(x)$.
Correct Option: 1Important Questions from Continuous Distributions
Suppose X is a continuous random variable with probability density function
\(f(x)=\frac{1}{\pi} \frac{1}{1+(x+1)^2}\), -∞ < x < ∞.
Define
\(Y=\left\{\begin{array}{cc} \frac{X}{|X|}, & \text { if } X \neq 0 \\ 0, & \text { if } X=0 \end{array}\right.\)
Then which of the following statements are true?
Let X1, X2, ..., Xn be a random sample from an absolutely continuous distribution with the probability density function
\(f(x \mid \theta)=\left\{\begin{array}{cl} e^{\theta-x}, & \text { if } x \geq \theta \\ 0, & \text { if } x<\theta \end{array},\right.\)
where θ ∈ ℝ is unknown. Define \(\bar{X}=\frac{1}{n} \sum_{i=1}^n X_i\) and X(1) = min{X1, ..., Xn}. Then
which of the following statements are true?
Suppose that X is a continuous random variable with probability density function given by:
f(x) = \(\left\{ {\begin{array}{c} {\frac{x}{8},}&{x \in \left[ {0,2} \right)}\\ {\frac{1}{4},}&{x \in \left[ {2,4} \right)}\\ { - \frac{x}{8} + \frac{3}{4},}&{x \in \left[ {4,6} \right)} \end{array}}\right.\)
Find the mean of X.
The variable x takes a value between 0 and 10 with uniform probability distribution. The variable y takes a value between 0 and 20 with uniform probability distribution. The probability of the sum of variables (x + y) being greater than 20 is _________
Probability density function of a random variable X is given below
\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {0.25}&{if\;1 \le x \le 5}\\ 0&{otherwise} \end{array}} \right.\)
P (X ≤ 4) is
