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If I have

$$f(x)=\sin(x\pi/10)\qquad\text{for}\;0\leq x\leq10.$$

How do I tell if it is a probability density function? And if it isn't how do I normalize it?

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You would have to calculate the net probability by integrating from $x=0$ to $x=10$. What does the "area" under a probability density function have to be? –  EuYu Oct 2 '12 at 15:52
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Also remember to check that $f(x)\geq 0$ for all $x$! –  Per Manne Oct 2 '12 at 15:58
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2 Answers 2

up vote 5 down vote accepted

The integral of a pdf must be equal to one:

$$\int_{-\infty}^{\infty} f(x) \, dx =1$$

In this case, since the function $g(x)=\sin(\pi x/10)$ is defined in $ 0\leq x \leq 10$ and $g(x) \geq 0$ in this interval:

$$\int_{-\infty}^\infty g(x) \, dx = \int_0^{10} \sin(\pi x /10)= \frac{20}{\pi}$$

Then, we scale function $g(x)$ with the inverse of this value and the fdp would be:

$$f(x)=\frac{\pi}{20} \sin \left( \frac{\pi x}{10} \right)$$

Hope this helps!

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You shouldn't have equality with f and the sine function is f is the pdf. Might also be good to point out that from 0 to 2 pi the sine is positive over half the cycle and negative over the other. So you need to check that you are only including the half cycle or part of the half cycle where the sine function is non negative. –  Michael Chernick Oct 2 '12 at 21:36
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You need to check two conditions,

1)- $ f(x) $ has to be non-negative,

2)- $\int_{-\infty}^{\infty} f(x) dx =1$.

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is this saying -f(x) or just f(x)? the hypen/minus_sign makes it ambigous.. –  naxa Jun 24 '13 at 12:02
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