Is this a special probability distribution?

Question

Does the distribution function:
$\frac{1}{\theta}e^\frac{-y}{\theta} $

Have a special name?

If not, how can I find the variance? I keep running into a dead end when I try.

Answer 1

As Alex said, your distribution is called exponential distribution with parameter $\frac{1}{\theta}$. The probability density function of an exponential distribution is usually written in terms of $x$ and a parameter $\lambda>0$, and is

$$f(x;\lambda)=\lambda e^{-\lambda x}\ $$ where $x\in [0,\infty]$.

So, as you see, letting $\lambda = \frac{1}{\theta}$ gives your distribution. To calculate the variance, let's first calculate the expected value $E[X]$:

$$E[X]=\int_0^{\infty}xf(x;\lambda)dx=\lambda \int_0^\infty xe^{-\lambda x}dx=\frac{\lambda}{\lambda^2}=\frac{1}{\lambda}$$ Then, denoting $\mu=\frac{1}{\lambda}$

$$Var[X]=\int_0^\infty (x-\mu)^2\lambda e^{-\lambda x}dx=\frac{\lambda}{\lambda^3}=\frac{1}{\lambda^2}$$

So, in your case, since $\frac{1}{\theta}=\lambda$, $$E[X]=\theta$$ and $$Var[X]=\theta^2$$