1

Given:

Let $X$ be exponential with parameter $\lambda$, that is

$$ f_X(x) = \begin{cases} \lambda e^{-\lambda x} & \text{if }x> 0, \\ 0 &\text{for }x\leq 0. \end{cases} $$

where $\lambda>0$ is called the rate of the distribution.

Question:

Show that this is a valid probability density function (pdf).

Attempt:

Need to show that $$ f(x) = \int^{\infty}_{-\infty} \lambda e^{-\lambda x}\, dx=1\\\\ $$ Thus, \begin{eqnarray*} f(x)&=&\int^0_{-\infty} \lambda e^{-\lambda x} \, dx+\int^\infty_0 \lambda e^{-\lambda x} \, dx\\[8pt] &=&0+\int^\infty_0 \lambda e^{-\lambda x} \, dx \\[8pt] &=&\lambda\int^\infty_0 e^{-\lambda x} \, dx \\[8pt] &=&1 \end{eqnarray*}

The problem is that my professor just told me that I skip a step right before I derive $1$. Any insight?

EDIT:

Given the great feedback, here is the answer:

\begin{eqnarray*} f(x)&=&\int^{\infty}_{-\infty} \lambda e^{-\lambda x} dx=1\\\\ f(x)&=&\int^{\infty}_{-\infty} f_X(x)dx\\ &=&\int^{0}_{-\infty} 0\,\,\, dx+\int^{\infty}_{0} \lambda e^{-\lambda x} dx\\ &=&0+\int^{\infty}_{0} \lambda e^{-\lambda x} dx\\ &=&\int^{\infty}_{0} \lambda e^{-\lambda x} dx\\ &=&\int^{\infty}_{0} e^{-\lambda x} \Big(\lambda \,\,\,dx\Big)\\ &=&\int^\infty_0e^{-u} du\\ &=&-e^{-u}\Big|^{\infty}_0\\ &=&1-\lim _{x \to \infty} e^{-u}\\ &=&1\\ \end{eqnarray*}

Olga
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  • Maybe you should add that $f_X$ is always non-negative and continuous except that in zero, because not every function in $L^1(\mathbb{R})$ with integral one is a pdf. – Jack D'Aurizio Sep 03 '14 at 22:08
  • Three answers have appeared so far, but I am still the only one who's up-voted the question. – Michael Hardy Sep 03 '14 at 22:53

3 Answers3

1

For more rigor, you could have written $$\begin{align} f(x) &=\int_{-\infty}^{\infty}f_X(x)\ dx\\ &=\int^{0}_{-\infty} 0\ dx+\int^{\infty}_{0} \lambda e^{-\lambda x} dx\\ &=-e^{-\lambda x} \Big|^{\infty}_{0} \\ &=1 \end{align}$$

1

There is an error in this line: $$ f(x)=\int^0_{-\infty} \lambda e^{-\lambda x} \, dx+\int^\infty_0 \lambda e^{-\lambda x} \, dx. $$ It should say $$ f(x)=\int^0_{-\infty} 0 \, dx+\int^\infty_0 \lambda e^{-\lambda x} \, dx. $$

Later, where you have $$ \int_0^\infty \lambda e^{-\lambda x}\,dx, $$ I would write $$ \int_0^\infty e^{-\lambda x} \Big(\lambda \,dx\Big) = \int_0^\infty e^{-u}\,du = \cdots[\text{fill in the blanks here}]\cdots =1. $$

Michael Hardy
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  • Hello Michael. My professor was talking about a new function that would serve as a replacement for some notations in the formula. I think you got this. – Olga Sep 03 '14 at 22:31
0

Your last equality is correct but to fast: $$\int_{0 }^\infty \lambda e^{-\lambda x}dx=\lambda\int_0^\infty e^{-\lambda x}=\lambda\left[-\frac{1}{\lambda}e^{-\lambda x}\right]_0^\infty =-\left[e^{-\lambda x}\right]_0^\infty =1-\lim_{x\to\infty }e^{-\lambda x}=1.$$

I think that your teacher just wanted you to explain this part.

idm
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