GATE Chemical 2025 | Question: 27

Question

​​​A probability distribution function is given as
$$p(x)=\left\{\begin{array}{ll} \frac{1}{a}, & x \in(0, a) \\ 0, & \text { otherwise } \end{array}\right.$$

where $a$ is a positive constant. For a function $f(x)=x^{2}$, the expectation of $f(x)$ is

• A. $\frac{a^{2}}{3}$
• B. $\frac{a^{3}}{3}$
• C. $\frac{2 a^{2}}{3}$
• D. $\frac{2 a^{3}}{3}$

Answer 1

In general $ E[X] = \int x f(x)dx $ where $x$ is the variable's value itself and $f(x)$ is the pdf.
Here the main initial difficulty (for me atleast) was wrapping my head around the notation.

Essentially $f(x)$ is our random variable, with $p(x)$ as its pdf and the domain of $p(x)$ as its domain.

 

Therefore 

$ E [X] = \int_0^a f(x) p(x)dx $ 

 

$ = \int_0^a x^2 \frac{1}{a}dx $

 

$= [^a_0 \frac{x^3}{3} \times \frac{1}{a} $

 

$= \frac{a^3}{3a} - \frac{0^3}{3a} $

 

$\large = \frac {a^2}{3}$