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

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

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

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- Sep 29th 2010, 03:46 AMnoob mathematicianWhat distribution is this?
Let $\displaystyle s(x)=exp(-(\frac{x}{2})^c)$

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

How does the distribution look like? Can I conclude that $\displaystyle s(x)$ takes the form of an exponential distribution? - Sep 29th 2010, 05:06 AMamul28yeh...
- Sep 29th 2010, 05:18 AMmr fantastic
Since there is no normalising constant (which will be $\displaystyle \frac{1}{\Gamma \left(\frac{c+1}{c}\right)}$, 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).

- Sep 29th 2010, 06:44 AMamul28
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. - Sep 29th 2010, 01:25 PMmr fantastic
- Sep 29th 2010, 10:34 PMnoob 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

- Sep 30th 2010, 03:27 AMmr fantastic
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