If we have a probability density function given by $f(y)=\frac{a}{y^2}$ where $0, how do we find F(y)?
Finding a CDF from a PDF
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probability
probability-distributions
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0$F(y)\neq 1-\frac{y}{a}$ but $1-\frac{a}{y}$ . – 2012-07-07
1 Answers
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The cumulative distribution function is defined as:
$F(y)=\int_{-\infty}^yf(u)\ du$
So for your probability density function:
$F(y)=\int_{a}^y\frac{a}{u^2}\ du=[-\frac{a}{u}]_a^y=1-\frac{a}{y}$
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0@johnnymath Shouldn't you be substituting limit values $0$ and $\infty$ in the transition from $\int_0^\infty \frac{1}{y^2}$ to $-\frac{a}{y}$? – 2012-02-28