What distribution is this?

**noob mathematician** · September 29th 2010, 03:46 AM

Let $s(x)=exp(-(\frac{x}{2})^c)$

$x\ge 0$ , $c>0$

How does the distribution look like? Can I conclude that $s(x)$ takes the form of an exponential distribution?

**amul28** · September 29th 2010, 05:06 AM

Originally Posted by noob mathematician

Let $s(x)=exp(-(\frac{x}{2})^c)$

$x\ge 0$ , $c>0$

How does the distribution look like? Can I conclude that $s(x)$ takes the form of an exponential distribution?

yes this takes form of exponential distribution.

**mr fantastic** · September 29th 2010, 05:18 AM

Originally Posted by noob mathematician

Let $s(x)=exp(-(\frac{x}{2})^c)$

$x\ge 0$ , $c>0$

How does the distribution look like? Can I conclude that $s(x)$ takes the form of an exponential distribution?

Since there is no normalising constant (which will be $\frac{1}{\Gamma \left(\frac{c+1}{c}\right)}$ , by the way) it does not define a pdf. Overlooking that, I don't think it has any special name. It certainly is NOT an exponential distribution. The closest I can think of is the exponential power distribution (except it's not).

**amul28** · September 29th 2010, 06:44 AM

O sorry but, y can't it be exponential ?
I thought it is exponential because, as C is a constant, its value can b found by integrating the fn' . Isn't this interpertation correct.

**mr fantastic** · September 29th 2010, 01:25 PM

Originally Posted by amul28

O sorry but, y can't it be exponential ?
I thought it is exponential because, as C is a constant, its value can b found by integrating the fn' . Isn't this interpertation correct.

IF c = 1 then the function, after being multiplied by an appropriate normalising constant, will define an exponential pdf.

**noob mathematician** · September 29th 2010, 10:34 PM

Actually was asking about how it will look like.. But i guess exponential will be the closer and given that x is only defined >0

**mr fantastic** · September 30th 2010, 03:27 AM

Originally Posted by noob mathematician

Actually was asking about how it will look like.. But i guess exponential will be the closer and given that x is only defined >0

Try different values of c eg. c = 1/2, 1, 2, 3. Plot the graphs. What do you find?

c = 1/2: plot y = Exp[-(x/2)^1/2] - Wolfram|Alpha

c = 1: plot y = Exp[-(x/2)^1] - Wolfram|Alpha

c = 2: plot y = Exp[-(x/2)^2] - Wolfram|Alpha

c = 3: plot y = Exp[-(x/2)^3] - Wolfram|Alpha

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