Question:
If a random variable $X$ has the p.d.f. $f(x) = \begin{cases} \frac{k}{x^2+1} & , \text{if } 0<x<\infty \\ 0 & , \text{otherwise} \end{cases}$ then c.d.f. of X is
If a random variable $X$ has the p.d.f. $f(x) = \begin{cases} \frac{k}{x^2+1} & , \text{if } 0<x<\infty \\ 0 & , \text{otherwise} \end{cases}$ then c.d.f. of X is
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Always normalize p.d.f. first before finding c.d.f.
Updated On: Apr 26, 2026
- $2 \tan^{-1} x$
- $\frac{\pi}{2} \tan^{-1} x$
- $\frac{2}{\pi} \tan^{-1} x$
- $\tan^{-1} x$
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The Correct Option is C
Solution and Explanation
Concept:
For a valid p.d.f., total probability must be 1: \[ \int_{0}^{\infty} f(x)\,dx = 1 \] Step 1: Find constant \(k\). \[ \int_0^\infty \frac{k}{x^2+1} dx = 1 \] \[ k \int_0^\infty \frac{dx}{x^2+1} = 1 \] \[ k \left[ \tan^{-1} x \right]_0^\infty = 1 \] \[ k \left(\frac{\pi}{2} - 0\right) = 1 \] \[ k = \frac{2}{\pi} \]
Step 2: Find c.d.f. \(F(x)\). \[ F(x) = \int_0^x \frac{2}{\pi} \cdot \frac{1}{1+t^2} dt \] \[ F(x) = \frac{2}{\pi} \tan^{-1} x \] (for \(x>0\))
Step 3: Conclusion. \[ {F(x) = \frac{2}{\pi} \tan^{-1} x} \]
For a valid p.d.f., total probability must be 1: \[ \int_{0}^{\infty} f(x)\,dx = 1 \] Step 1: Find constant \(k\). \[ \int_0^\infty \frac{k}{x^2+1} dx = 1 \] \[ k \int_0^\infty \frac{dx}{x^2+1} = 1 \] \[ k \left[ \tan^{-1} x \right]_0^\infty = 1 \] \[ k \left(\frac{\pi}{2} - 0\right) = 1 \] \[ k = \frac{2}{\pi} \]
Step 2: Find c.d.f. \(F(x)\). \[ F(x) = \int_0^x \frac{2}{\pi} \cdot \frac{1}{1+t^2} dt \] \[ F(x) = \frac{2}{\pi} \tan^{-1} x \] (for \(x>0\))
Step 3: Conclusion. \[ {F(x) = \frac{2}{\pi} \tan^{-1} x} \]
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