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If we have a probability density function given by $f(y)=\frac{a}{y^2}$ where $0

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    The 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 $y2012-02-28
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    $F(y)\neq 1-\frac{y}{a}$ but $1-\frac{a}{y}$ .2012-07-07

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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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    ...if $y\geqslant a$ and $F(y)=0$ otherwise.2012-02-28
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    @DidierPiau Yeah,you're right!2012-02-28
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    Thanks! 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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    @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