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
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
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 _________
A nationalized bank has found that the daily balance available in its savings accounts follows a normal distribution with a mean of Rs. 500 and a standard deviation of Rs. 50. The percentage of savings account holders, who maintain an average daily balance more than Rs 500 is _______
The number of parameters in the univariate exponential and Gaussian distributions, respectively are
Find the value of λ such that the function f (x) is a valid probability density function. _______
\(f\left( x \right)\begin{array}{*{20}{c}} { = \lambda \left( {x - 1} \right)\left( {2 - x} \right)}&{for1 \le x \le 2}\\ { = 0}&{otherwise} \end{array}\)
