What is the Optimal Value of k for a Probability Density Function?

[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)

 

[radou]

What are the properties of a probability density function?

 

[theperthvan]

Area under curve = 1

 

[radou]

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.

 

[theperthvan]

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

 

[D H]

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?

 

[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.