Determine F(3) and F(1/$\lambda$) from a frequency function

Question

I have this problem:

The life span $\xi$ in a radioactive atom has the frequency function:

$$f(x)=\begin{cases}\lambda e^{-\lambda x}&x>0\\ 0&x < 0\end{cases}$$

Determine $F(3)$ and $F(1/\lambda)$

I don't know how to do this. It seems like it creates an integral going from 0 to infinity but I got stuck there.

The answer is F(3) = $1-e^{-3\lambda}$ and F($1/\lambda$) = 0.63

Answer 1

I assume in the above that you are using the term frequency function to describe a probability density function, and that $F$ is the associated cumulative distribution function.

In this case

\begin{align*} F(x) & =\mathbf{P}[ \xi \leq x] \\ & = \int_{0}^x \lambda e^{-\lambda y} dy \\ & = \lambda \left[ -\frac{1}{\lambda}e^{-\lambda y} \right]_{0}^x \\ & = 1 - e^{-\lambda x}. \end{align*}

Hence we have

$$F(3) = 1 - e^{-3 \lambda}$$ whilst $$F(1/\lambda) = 1 - e^{-1} \approx 0.63$$

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Computing the anti-derivative of $e^{\alpha x}$

In the above, I assumed the indefinite integral formula

$$ \int e^{\alpha y} d y = \frac1\alpha e^{\alpha y},$$

for general $\alpha \in \mathbf{R}$. Probably the easiest way to justify this is by arguing that integration is the `opposite' of differentiation. If you are happy with the formula that $$ \frac{d}{dx} e^{\alpha x} = \alpha e^{\alpha x}$$ then (very informally) $$ \int \alpha e^{\alpha x} = \int \frac{d}{dx} e^{\alpha x} = C + e^{\alpha x},$$ justifying the formula for the indefinite integral. To make this argument formal you would need to use the Fundamental Theorem of Calculus.

Answer 2

Note that in this case $$ F(x)=P(X\leq x)=\int_{0}^x f(t) \,dt. $$ In particular $$ F(3)=\int_{0}^3\lambda e^{-\lambda x}\, dx. $$ Hopefully you can compute this integral.