# A probability distribution

1. Jan 29, 2007

### theperthvan

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)

Last edited: Jan 29, 2007
2. Jan 29, 2007

What are the properties of a probability density function?

3. Jan 29, 2007

### theperthvan

Area under curve = 1

4. Jan 29, 2007

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

5. Jan 29, 2007

### mathman

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

6. Jan 29, 2007

### theperthvan

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

7. Jan 29, 2007

### D H

Staff Emeritus
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?

8. Jan 29, 2007

### theperthvan

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.