# Probability equation in case of an exponential distribution

• I
Gold Member

## Main Question or Discussion Point

Why is ##P(X>5|X>1) = P(X>4)## in case of an exponential distribution?

Can anyone kindly explain it to me?

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Math_QED
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Just write out the densities probabilities using density of exponential distribution and it will become clear.

Gold Member
Just write out the densities probabilities using density of exponential distribution and it will become clear.
I couldn't understand.

Math_QED
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I couldn't understand.

Gold Member
The density function for an exponential distribution is ##f(x; \lambda) = \lambda e ^{-\lambda x}## in case of ##x\ge0##.

The plot would be something like the following: Math_QED
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Okay, good. Do you know that ##P(X \leq t) = \int_{0}^t \lambda e^{-\lambda x} dx## for ##t > 0##? If so, why don't you just calculate all probabilities involved and see that they are equal?

Gold Member
Okay, good. Do you know that ##P(X \leq t) = \int_{0}^t \lambda e^{-\lambda x} dx## for ##t > 0##? If so, why don't you just calculate all probabilities involved and see that they are equal?
1. Yes, I know.
2. I want need to understand it either graphically or by proof in order to solve understand the following problem as I already have the solution with me: Ray Vickson
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1. Yes, I know.
2. I want need to understand it either graphically or by proof in order to solve understand the following problem as I already have the solution with me:

View attachment 241847
I don't understand your lack of understanding. You have formulas for P(X>5|X>1)---just the ordinary conditional probability formula. So, if you know how to compute the numerator P(X > 5 & X > 1) you are done.

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I don't understand your lack of understanding. You have formulas for P(X>5|X>1)---just the ordinary conditional probability formula. So, if you know how to compute the numerator P(X > 5 & X > 1) you are done.
##P(X>5|X>1) = \frac{P(X>5, X>1)}{P(X>1)}##

Is this the correct formula?

Do not forget exponential distribution has a memoryless property. When you set a condition, you do not have to consider the cases before it.

Gold Member
Do not forget exponential distribution has a memoryless property. When you set a condition, you do not have to consider the cases before it.
So, can I write ##P(X>5|X>1) = P(X>5)## ?

Ray Vickson
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##P(X>5|X>1) = \frac{P(X>5, X>1)}{P(X>1)}##

Is this the correct formula?
What do YOU think? What is stopping you from carrying on, to complete the task?

• Math_QED
Gold Member
What do YOU think? What is stopping you from carrying on, to complete the task?
As I have already said, I have calculations with me. I am trying to understand the core idea.

If exponential distribution is memoryless (i.e. the past has no bearing on its future behavior), why can't I write ##P(X>5|X>1) = P(X>5)##?

• Periwinkle
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This doesn't answer my specific query.

Ray Vickson
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So, can I write ##P(X>5|X>1) = P(X>5)## ?
No, obviously not: the left-hand side evaluates to ##P(X > 4),## and that is different from ##P(X > 5)##.

Look at it intuitively: once you have reached time ##1## you have 4 units left to go to reach time ##5##. The memoryless property says that the probability of having to wait at least 4 is exactly the same as for a brand-new random variable X that that has just been born at time 0----in other words, you have the same P(X > 4) whether you have just started or have already waited 1 unit (or 50 units or ...... ).

The easiest way to write the memoryless property is to say that
$$P(X > t+s | X > t) = P(X > s)$$ So, if ##t=1## and ##t+s=5## we have ##s=4.##

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As I have already said, I have calculations with me. I am trying to understand the core idea.

If exponential distribution is memoryless (i.e. the past has no bearing on its future behavior), why can't I write ##P(X>5|X>1) = P(X>5)##?
I'll describe what I think of this. Probability theory does not generally deal with the calculation of probabilities, but with an extraordinary task. That's it
1. examining the laws of random mass phenomena,
2. under precisely defined experimental conditions.
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Suppose that the expected lifetime of a radioactive particle is ##1/\lambda##. In this case (I hope I remember it well) the distribution of its lifetime is exponential. The lack-of-memory property means that the particle is not aging, so if the particle has lived for a hundred years, it will have the same probability of surviving another hundred years as if it had just emerged in an atomic process.

The question, however, was whether the probability that the particle lifetime would be longer than five years provided that it had already lived for more than one year would be the same as the probability that the initially generated particle would last for more than five years. He forgets his past - that he already existed for a year - so it might seem right. On the other hand, the statement is false.

Calculate the probabilities.
$$P(X>5) = e^{-5\lambda},$$
$$P(X>5 | X>1) = \frac {P(X>5~ and~ X>1)} { P(X>1)} = \frac {P(X>5)}{ P(X>1)} = e^{-5\lambda}/e^{-\lambda} = e^{-4\lambda}.$$
The two quantities are different.

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However, as I wrote above, probabilities can only be discussed under predetermined experimental conditions.
If the experimental conditions are fixed so that only the random events in which the particle has lived for more than one year are tested, the same probability is that it will live for more than five years, provided it lived for more than one year, as the probability that the newly formed particle will live longer than five years. However, if we examine not only the particles that live longer than one year, but all the new particles, then these two quantities are different. The above calculation was for this.

The lack-of-memory property doesn't mean we forget what the experimental conditions are, which we are investigating.

Gold Member
I'll describe what I think of this. Probability theory does not generally deal with the calculation of probabilities, but with an extraordinary task. That's it
1. examining the laws of random mass phenomena,
2. under precisely defined experimental conditions.
................................................................................................................................................................................................
Suppose that the expected lifetime of a radioactive particle is ##1/\lambda##. In this case (I hope I remember it well) the distribution of its lifetime is exponential. The lack-of-memory property means that the particle is not aging, so if the particle has lived for a hundred years, it will have the same probability of surviving another hundred years as if it had just emerged in an atomic process.

The question, however, was whether the probability that the particle lifetime would be longer than five years provided that it had already lived for more than one year would be the same as the probability that the initially generated particle would last for more than five years. He forgets his past - that he already existed for a year - so it might seem right. On the other hand, the statement is false.

Calculate the probabilities.
$$P(X>5) = e^{-5\lambda},$$
$$P(X>5 | X>1) = \frac {P(X>5~ and~ X>1)} { P(X>1)} = \frac {P(X>5)}{ P(X>1)} = e^{-5\lambda}/e^{-\lambda} = e^{-4\lambda}.$$
The two quantities are different.

................................................................................................................................................................................................
However, as I wrote above, probabilities can only be discussed under predetermined experimental conditions.
If the experimental conditions are fixed so that only the random events in which the particle has lived for more than one year are tested, the same probability is that it will live for more than five years, provided it lived for more than one year, as the probability that the newly formed particle will live longer than five years. However, if we examine not only the particles that live longer than one year, but all the new particles, then these two quantities are different. The above calculation was for this.

The lack-of-memory property doesn't mean we forget what the experimental conditions are, which we are investigating.
@Periwinkle Thanks for the answer! That concludes my query.