Distribution for this Probability density function?

Question

I having problem recognizing the following distribution. The random variable has density

$$f_\theta(x)=\frac{x}{\theta}\exp(-x^2/ 2\theta)1_{(0,\infty)}(x)$$

with respect to lebesgue, with parameter $\theta>0$.

Edit: A few more questions :D

When we talk about distribution of a random variable, there are a number of ways you can specify it right? In this case, to say that $X\sim R(\theta)$ is equivalent to saying $X$ has probability measure $\nu=f_{\theta}\cdot m$ ?

furthermore, how can I find the distribution of $X^2$?

The Rayleigh distribution is your friend. – wolfies Mar 21 '15 at 17:57 — wolfies, Mar 21 '15 at 17:57

score 1 · Accepted Answer · edited Jun 12 '20 at 10:38

1

This is the Rayleigh distribution with parameter $θ$ instead of $σ^2$ (as given in Wikipedia):

The probability density function of the Rayleigh distribution is

$$f(x;\sigma) = \frac{x}{\sigma^2} e^{-x^2/(2\sigma^2)}, \quad x \geq 0$$

where $\sigma$ is the scale parameter of the distribution. The cumulative distribution function is $$F(x) = 1 - e^{-x^2/(2\sigma^2)}$$ for $x \in [0,\infty).$

edited Jun 12 '20 at 10:38

Community

1

answered Mar 21 '15 at 17:57

Jimmy R.

35,868

Have a follow up question in my edit if you don't mind. – ChuckP Mar 21 '15 at 18:28
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@ChuckP The probability that $X^2 > y$ is the probability that $X> \sqrt{y}$, since all is positive. – Karl Mar 21 '15 at 18:46
I see now, it gives me exp dist with 1/2\theta as rate parameter. – ChuckP Mar 21 '15 at 19:17

Distribution for this Probability density function?

1 Answers1