What are the properties of this cumulative density function

Question

I'm given a probability density function $\lambda e^{-\lambda x}$, I therefore deduce that the cumulative density function is its ingegral: $\int -exp(-x\lambda)$

I'm trying to find out the properties of this density function. When I look at the plot in wolfram alpha I come to the conclusion that it needs to have a $\lambda=i (π + 2 n π)$

Is this a valid pdf?

Answer 1

The support of an exponential distribution is the non-negative real numbers.

Your cumulative distribution function should therefore be $\mathbb P(X \le x)= \int\limits_0^x \lambda e^{-\lambda t} \,dt = 1 - e^{-\lambda x} $ for $x\ge 0$

The CDF is then an increasing function, taking the value $0$ when $x=0$ and approaching $1$ when $x$ increases without limit. By contrast, the PDF (probability density function) $\lambda e^{-\lambda t}$ is a decreasing function on the non-negative reals taking the value $\lambda$ when $t=0$ and approaching $0$ when $t$ increases without limit

There is no need to consider complex numbers

Answer 2

It would behoove you to think more carefully about the definitions of the mathematical objects you are considering, instead of throwing it into Wolfram Alpha with incorrect syntax and blindly interpreting the output.

One thing you are missing from the given probability density function is the domain. The function is $f(t) = \lambda e^{-\lambda t}$ for $t > 0$; otherwise $f(t) = 0$. This is quite important when you go to compute the cumulative density function.

The cumulative density function of a random variable with density $f$ is indeed an integral of $f$, but the limits of integration are important. $$F(t) = \int_{-\infty}^t f(x) \, dx.$$ Note that the argument of the function $F$ is the upper limit of the integral. In the case of your particular density, $$F(t) = \int_0^t \lambda e^{-\lambda x} \, dx.$$