If we have a probability density function given by $f(y)=\frac{a}{y^2}$ where $0
Finding a CDF from a PDF
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probability
probability-distributions
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0The cumulative distribution function is in principle defined for all $y$. So a complete answer would be $F(y)=0$ if $y $F(y)=1-\frac{y}{a}$ if $y \ge a$. Depending on the mood of the grader, leaving out the uninteresting part $F(y)=0$ if $y – 2012-02-28
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0$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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1...if $y\geqslant a$ and $F(y)=0$ otherwise. – 2012-02-28
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0@DidierPiau Yeah,you're right! – 2012-02-28
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0Thanks! But I don't understand how $F(y)=1-\frac{a}{y}$. The way I did it originally, (but it didn't make sense, which is why this question is up) I had $\int_0^\infty \frac{a}{y^2}=\frac{-a}{y}$ – 2012-02-28
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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