verify that the following are probability density function.
F(x)=1/π 1/π(1+x^2),-∞‹x‹∞
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Probability density function has a value between 0 and 1. The (sum of) total value of probability for all values in the domain of x, is always 1. So we need to check that.
f(x) = 1/π * 1/(1+x²) , -∞ < x < ∞
1+x² > 1. Hence f(x) < 1 for x
f(x) is defined for each value of x in the domain.
Cumulative probability function:
![F(x)= \int \limits_{-\infty}^{x} {f(x)} \, dx\\\\=\frac{1}{\pi}\int \limits_{-\infty}^{x} {\frac{1}{1+x^2}} \, dx\\\\=\frac{1}{\pi} [ tan^{-1} x]_{-\infty}^{x}\\\\=\frac{1}{\pi} [ Tan^{-1} x+\frac{\pi}{2}]\\\\So \ F(-\infty)=0, \ F(0)=\frac{1}{2}, \ F(\infty)=1\\](https://tex.z-dn.net/?f=F%28x%29%3D+%5Cint+%5Climits_%7B-%5Cinfty%7D%5E%7Bx%7D+%7Bf%28x%29%7D+%5C%2C+dx%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B%5Cpi%7D%5Cint+%5Climits_%7B-%5Cinfty%7D%5E%7Bx%7D+%7B%5Cfrac%7B1%7D%7B1%2Bx%5E2%7D%7D+%5C%2C+dx%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B%5Cpi%7D+%5B+tan%5E%7B-1%7D+x%5D_%7B-%5Cinfty%7D%5E%7Bx%7D%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B%5Cpi%7D+%5B+Tan%5E%7B-1%7D+x%2B%5Cfrac%7B%5Cpi%7D%7B2%7D%5D%5C%5C%5C%5CSo+%5C+F%28-%5Cinfty%29%3D0%2C+%5C+F%280%29%3D%5Cfrac%7B1%7D%7B2%7D%2C+%5C+F%28%5Cinfty%29%3D1%5C%5C "F(x)= \int \limits_{-\infty}^{x} {f(x)} \, dx\\\\=\frac{1}{\pi}\int \limits_{-\infty}^{x} {\frac{1}{1+x^2}} \, dx\\\\=\frac{1}{\pi} [ tan^{-1} x]_{-\infty}^{x}\\\\=\frac{1}{\pi} [ Tan^{-1} x+\frac{\pi}{2}]\\\\So \ F(-\infty)=0, \ F(0)=\frac{1}{2}, \ F(\infty)=1\\")
Hence, f(x) is a probability density function.
The mean μ of the distribution is x=0, as cumulative probability is 1/2 for x=0.
f(x) = 1/π * 1/(1+x²) , -∞ < x < ∞
1+x² > 1. Hence f(x) < 1 for x
f(x) is defined for each value of x in the domain.
Cumulative probability function:
Hence, f(x) is a probability density function.
The mean μ of the distribution is x=0, as cumulative probability is 1/2 for x=0.
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