Prove $F(x)=1-e^{- \lambda x}, 0 \lt x.$

Question

Let $F$ be the distribution function of the continuous-type random variable $X$, and assume that $F(x)=0$ for $x \le 0$ and $0 \lt F(x) \lt1$ for $0 \lt x$. Prove that if $P(X \gt x+y |X \gt x)=P(X \gt y),$ then $F(x)=1-e^{- \lambda x}, 0 \lt x.$

My attempt:

Let $F(x)=P(X \le x)$. Then

$ \frac {P(X \gt x+y)}{P(X \gt x)}=P(X \gt y)$

$\Rightarrow \frac {1-P(X\le x+y)}{1-P(X \le x)}=1-p(X \le y)$

$ \Rightarrow \frac {1-F(x+y)}{1-F(x)}=1-F(y)$

Then I have no idea where to go from here. Any help is appreciated.

score 1 · Answer 1 · answered Jun 24 '17 at 03:24

1

Let $F(s) = P(X>s)$ Using conditional probability manipulations, $$F(s) = P(X>s) = P(X > s+t\mid X >t) = \frac{P(X>s+t)}{P(X>t)} = \frac{F(s+t)}{f(t)}$$ yielding the functional equation $$F(s+t) = F(s)F(t)$$ Next we consider the difference quotient $$\frac{F(x+h)-F(x)}{h} = F(x)\left(\frac{F(h)+1}{h}\right)$$ Taking limits of both sides and applying the definition of the derivative, we obtain the differential equation $$F'(x) = -\lambda F(x)$$ where $\lambda = -F'(0)$. The solution of this differential equation is the exponential function, $F(x) = e^{-\lambda x}$. The cumulative distribution function is then $$P(X\le x) = 1-P(X>x) = 1-e^{-\lambda x}$$

answered Jun 24 '17 at 03:24

Jonathan Davidson

788

Are you saying that $ \frac {F(h)+1}{h}=\lambda$? – Lily Jun 24 '17 at 15:52
I am saying the $\lim_{h \to 0}\frac{F(h)+1}{h} = F'(0) = \lambda$ – Jonathan Davidson Jun 24 '17 at 17:54
I am lost after you consider the difference quotient. Could you add more details of how you concluded that $\lambda=-F'(0)$ and how you have concluded that it is exponential function, $F(x)=e^{-\lambda x}$. Thank you. – Lily Jun 24 '17 at 21:46
$F'(0)$ is just a constant, so i named it $\lambda$ – Jonathan Davidson Jun 24 '17 at 21:54
I got that part now. How did you know from there to conclude that $F(x)=e^{- \lambda x}$? – Lily Jun 24 '17 at 22:40
I solved the differntial equation $F' = -\lambda F$ – Jonathan Davidson Jun 24 '17 at 22:41
If I differentiate $-\lambda F$, then it equals to $- \lambda$, right? – Lily Jun 24 '17 at 22:56
No. You do not solve a differential equation by differentiating, you do it by integrating – Jonathan Davidson Jun 24 '17 at 22:58
Let us continue this discussion in chat. – Jonathan Davidson Jun 24 '17 at 22:58

score 0 · Answer 2 · answered Jun 24 '17 at 03:23

0

Hint: let $y \to 0$ and try to get a differential equation for $F(x)$

answered Jun 24 '17 at 03:23

leonbloy

66,202

Prove $F(x)=1-e^{- \lambda x}, 0 \lt x.$

2 Answers2