# A probability distribution

## Main Question or Discussion Point

Here is a question that was in on of my exams a few months ago. It asks what is the value of k so that the function f(x;k) is a probability density.
I didn't really answer it, but put an answer as $$k= \frac{1}{2\sqrt{\pi}}$$ because that somewhat resembled the Normal distribution.
Does anyone know?

$$f(x;k) = -(\frac{1}{2\sqrt{\pi}} - k)^2 + k.e^{(10x - 0.25x^2 - 100)}$$

(this is not a homework question)

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Homework Helper
What are the properties of a probability density function?

Area under curve = 1

Homework Helper
Area under curve = 1
Right, i.e. $$\int_{-\infty}^{+\infty} f(x)dx =1$$

mathman
You also need f(x)>=0 for all x for f(x) to be a probability density.

But $$k.e^{(10x - 0.25x^2 - 100)}$$ isn't able to be integrated using elementary functions.

D H
Staff Emeritus
But $$k.e^{(10x - 0.25x^2 - 100)}$$ isn't able to be integrated using elementary functions.
That doesn't matter, does it? The standard normal distribution integrates to 1:

$$\int_{-\infty}^{\infty} \frac 1 {\sqrt{2\pi}} \exp\left(-\,\frac1 2 x^2\right) = 1$$

From this, you should be able to calculate the integral of $k.e^{(10x - 0.25x^2 - 100)}$ over all x.

BTW, your guess was right because setting $k= \frac{1}{2\sqrt{\pi}}$ eliminates the first term and makes the latter term a normal PDF. Can you see why?

Ahh, right. I guessed it for that reason (to eliminate the other term), but kind of by accident.

I see how it goes now. Thanks for your responses.