MHB Exponential distribution question

AI Thread Summary
The discussion centers on understanding the exponential distribution and its related functions, specifically \(R_X(y)\), \(R'_X(y)\), and \(F'_X(0)\). It clarifies that \(R_X(y)\) is defined as \(1 - F_X(y)\), leading to the relationship \(R_X'(y) = -F_X'(y)\) and confirming \(R_X(0) = 1\). The limit process used to derive these functions is explained, emphasizing the connection between the cumulative distribution function (CDF) and the probability density function (PDF). The solution to the differential equation is confirmed as \(R_X(y) = K \cdot e^{R'_X(0) y}\). This discussion highlights the mathematical relationships within the exponential distribution framework.
WMDhamnekar
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Hi,
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I want to know how the highlighted steps are arrived at in the first page. What are \(R_X (y), R'_X (y),F'_X (0) ? \)How \(R_X (0) = 1 ?\) Solution to differential equation should be \(R_X (y)=K*e^{\int{R'_X (0) dx}}\) But it is different. How is that?

What is $-R'_X (0)=F'_X(0)=f_X(0)$ I know derivative of CDF is PDF, but in this case it is somewhat difficult to understand.

If any member of MHB knows how to satisfy my queries correctly, may reply to this question:confused::unsure:
 
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Hey Dhamnekar Winod,

In the first step they apply:
$$\lim_{t\to 0} \frac{F_X(t)[1-F_X(y)]}{t} = \lim_{t\to 0} \frac{F_X(t)-F_X(0)}{t}\cdot [1-F_X(y)] = F_X'(0)\cdot [1-F_X(y)]$$

Apparently they defined $R_X(y)=1-F_X(y)$ as an intermediate step to solve the differential equation.
It follows that $R_X'(y)=-F_X'(y)$ and $R_X(0)=1$.
The corresponding equation then follows from the previous equation.

The solution of the differential equation is indeed:
$$R_X (y)=K\cdot e^{\displaystyle\int_0^y{R'_X (0) dx}} = K\cdot e^{\displaystyle\big[R'_X (0) x\big]_0^y} = K\cdot e^{R'_X (0) y}$$
 
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