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The exercise

A continuous random variable X, expresses in minutes, the duration of the telephone communications. Its density function is as follows: $f(x)=ke^{-\alpha x}$ if $x>0$, else $f(x)=0$

  • a) What conditions must the real $α$ and $k$ verify to be a density function?

If $k=5/6$:

  • b) Get the distribution function of X.
  • c) Calculates the probability that a communication will last more than three minutes.
  • d) Calculates the probability that a communication will last more than five minutes, if it is known to have lasted more than two minutes.

My doubts

  • I know what conditions have to meet a density function. But in part (a), how should I analyze these two real numbers? Is it with limits? If so, how should I evaluate them?
  • In part (d), should I interpret it as a conditional probability? If so, in what way can I calculate it?

If you have more to say about the rest of the exercise, you are also welcome. Thank you very much.

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    The conditions that a density function have to meet will give you restrictions on $k$ and $\alpha$. For (d), you can interpret it as a conditional probability, for which you'll need to derive a conditional density. You might discover an interesting property of the exponential distribution in the process. – 2017-02-14
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    @Theoretical Economist Yes, but part (a) I'm doing it by observation (using my calculator). I wanted to know if it could be done in a more direct way. In part (d), could you explain a little more about the conditional probability for this point case? – 2017-02-14

1 Answers 1

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(a) You want, $f(x)\geq 0$ for all $x$, and $\int_\Bbb R f(x)\operatorname d x=1$.   So what constraints on $k, \alpha$ make this possible?

(b) $\mathsf P(X\leq b) = \int_{-\infty}^b f(x)\operatorname d x$

(c) $\mathsf P(X>b) = \int_b^\infty f(x)\operatorname d x$

(d) You want $\mathsf P(X > 5\mid X > 2)$, so use Bayes' Rule and the above.

Tip: $\int_b^c e^{-ax} \operatorname d x = \tfrac 1a(e^{-ab}-e^{-ac})$

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    I did the calculations and they gave me $\displaystyle\alpha=\dfrac{5}{6}$ and $\displaystyle F_X(x)=1-e^{-\dfrac{5}{6}x}$. Is there any way to verify if what I did was okay? – 2017-02-14
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    @AccomOuter Verified: What you did was okay. – 2017-02-14