1. ## Exponential Dist

Hi guys,

A sample of 5 iid is taken from an exponential distribution with λ =1.
1) What is the P(smallest observation is < 0.5)?
2) What is the P(largest observation is < 0.5)?

I am thrown off by the smallest/largest condition.

Many thanks in advance for any feedback.
Linda

2. P(smallest observation is < 0.5) = p(at least one < 0.5) = 1 - P(all ovservations >= 0.5)

can you evaluate the RHS?

The other question is similar

3. Is it, using the cdf:

=1 - e^{\lambda x}
=1 - e^{1 x 0.5}
=0.3934

With the largest observation, would this be:
P(largest observation is < 0.5) = 1 - P(all ovservations >= 0.5)
because of the fact that the exponential distribution is memoryless??

4. I did not check your arithmatic. Assuming you calculated properly 0.3934 would be the chance of one obvservation being greater than 0.5.

the chance of 5 being great than 0.5 would be 0.3934 ^ 5

P(largest observation is < 0.5) = 1 - P(all ovservations >= 0.5)
because of the fact that the exponential distribution is memoryless??
I dont think that's right.

p(largest < 0.5) = p(all are < 0.5)

This would be true for any distribution. I dont think the memorylessness property is relevent here, but i could be wrong.

5. i claim everything i say is inaccurate, but then this is inaccurate

6. Originally Posted by matheagle
i claim everything i say is inaccurate, but then this is inaccurate
Mine too!!!
Is it possible for you to elaborate?

Originally Posted by SpringFan25
I did not check your arithmetic. Assuming you calculated properly 0.3934 would be the chance of one observation being greater than 0.5.
I know you have not checked the arithmetic - the chance of one observation greater than 0.5 is 0.6065.
So I am gathering the answer is

p(largest < 0.5) = p(all are < 0.5)

This would be true for any distribution. I don't think the memoryless property is relevant here, but i could be wrong.
If that is the case would this calculation then be:
=(1-0.6065)^5