Exponential Distribution memory loss

[Mentallic]

Homework Statement

Show that the exponential distribution has the memory loss property.

Homework Equations

[tex]f_T(t) = \frac{1}{\beta}e^{-t/\beta}[/tex]

The memory loss property exists if we can show that

[tex]P(X>s_1+s_2|X>s_1) = P(X>s_2)[/tex]

Where

[tex]P(X>s_1+s_2|X>s_1)=\frac{P(X>s_1+s_2,X>s_1)}{P(X>s_1)}[/tex]

The Attempt at a Solution

For [itex]P(X>s_2)[/itex] I computed the integral

[tex]1-\int_0^{s_2}f_T(t)dt[/tex]

And similarly I found [itex]P(X>s_1)[/itex] but I'm unsure how to find [itex]P(X>s_1+s_2,X>s_1)[/itex] to complete the problem.

 

[HallsofIvy]

Homework Statement

Show that the exponential distribution has the memory loss property.

Homework Equations

[tex]f_T(t) = \frac{1}{\beta}e^{-t/\beta}[/tex]

The memory loss property exists if we can show that

[tex]P(X>s_1+s_2|X>s_1) = P(X>s_2)[/tex]

Where

[tex]P(X>s_1+s_2|X>s_1)=\frac{P(X>s_1+s_2,X>s_1)}{P(X>s_1)}[/tex]

This is miswritten- probably a typo. You mean
[tex]P(X> x_1+ s_2|X> s_1)= \frac{P(X> s_1+ s_2)}{P(X> s_1)}[/tex]
That is, the numerator on the right is NOT the condition probability (if it were you could cancel it with the left).

The Attempt at a Solution

For [itex]P(X>s_2)[/itex] I computed the integral

[tex]1-\int_0^{s_2}f_T(t)dt[/tex]

And similarly I found [itex]P(X>s_1)[/itex] but I'm unsure how to find [itex]P(X>s_1+s_2,X>s_1)[/itex] to complete the problem.

That would be
[tex]\frac{1- \beta\int_0^{s_1+ s_2}e^{-t/\beta} dt}{1- \beta\int_0^{s_1} e^{-t/\beta}dt}[/tex]

 

[Ray Vickson]

Homework Statement

Show that the exponential distribution has the memory loss property.

Homework Equations

[tex]f_T(t) = \frac{1}{\beta}e^{-t/\beta}[/tex]

The memory loss property exists if we can show that

[tex]P(X>s_1+s_2|X>s_1) = P(X>s_2)[/tex]

Where

[tex]P(X>s_1+s_2|X>s_1)=\frac{P(X>s_1+s_2,X>s_1)}{P(X>s_1)}[/tex]

The Attempt at a Solution

For [itex]P(X>s_2)[/itex] I computed the integral

[tex]1-\int_0^{s_2}f_T(t)dt[/tex]

And similarly I found [itex]P(X>s_1)[/itex] but I'm unsure how to find [itex]P(X>s_1+s_2,X>s_1)[/itex] to complete the problem.

This property is normally called memoryless, not memory loss. Anyway, if ##s_1, s_2 > 0## then ##P(X > s_1+s_2, X > s_1) = P(X > s_1 + s_2),## because if ##X > s_1 + s_2## then we automatically have ##X > s_1##.

 

[Mentallic]

This is miswritten- probably a typo. You mean
[tex]P(X> x_1+ s_2|X> s_1)= \frac{P(X> s_1+ s_2)}{P(X> s_1)}[/tex]
That is, the numerator on the right is NOT the condition probability (if it were you could cancel it with the left).

Actually I had it right the first time, but I didn't quite understand what it meant. I now realize that my notes' mention of

[tex]P(X>s_1+s_2,X>s_1)[/tex]

is equivalent to

[tex]P(X>s_1+s_2 \cap X>s_1)[/tex]

and Ray supported that claim:

Anyway, if ##s_1, s_2 > 0## then ##P(X > s_1+s_2, X > s_1) = P(X > s_1 + s_2),## because if ##X > s_1 + s_2## then we automatically have ##X > s_1##.

Thanks guys!

This property is normally called memoryless, not memory loss.

Yeah, that's my fault