# Expectation Of The Maximum When One Of The Random Variables Is Constant

1. Jul 17, 2009

### actcs

Good Evening:

I'm given this problem:

A device that continuously measures and records seismic activity is placed in a remote
region. The time, T, to failure of this device is exponentially distributed with mean
3 years. Since the device will not be monitored during its first two years of service, the
time to discovery of its failure is X = max(T, 2) .
Determine E[X].

Solution: 2 + 3 Exp[-2/3]

I've even got the procedure

It's:

E[X] = Integral From 0 To 2 [ 2*f(t)dt ] + Integral From 2 To Infinity [ t*f(t)dt ]

Where f(t)=1/3 Exp[-1/3 t]

I Just want to know, why is this?... Why the interval of the first integral is from 0 To 2, and then again that "2" appears in the integral?... I tried to calculate by means of order statistics but result didn't match

Does someone know how to prove this is actually the solution?... (I'm certain this is the correct solution, but I just want a more specific and justified procedure. I'm not familiarized with constant random variables)

2. Jul 17, 2009

### tiny-tim

Good morning, actcs!

Because E[X] = ∫ X(t) f(t) dt.

So when X(t) is a constant, K, over an interval, the integral over that interval is ∫ K f(t) dt …

and in this case K = 2.

(and ∫ f(t) dt, without the K, would just be the probability, not the expectation)

3. Jul 17, 2009

### actcs

Hello, Thank you for replying

The problem in this case is that the constant random variable is involved in an order statistic, so it is not so trivial to see the sample space of each of the random variables

I did this: Write the random variable X as:

X = 2 I(0,2](T) + T I(2,Infinity)(T)

Where I(a,b)(T) is an indicator function for the random variable T

The way I see this was:

If device fails between now and the end second year, discovery time will be "End of year 2", whereas if device fails after that, discovery time will match fail time, ie, X=T

I think it was rather complicated to obtain the density funcion of X and then calculate the expectation

Best Regards

4. Jul 17, 2009

### John Creighto

I don't think that is the correct solution since:

Integral from 0 to 2 of Exp[-1/3 t]= -Exp[-1/3 t]-(1)=1-Exp[-1/3 t]

Therefore your solution should be:

E[X] = 2 [1-Exp[-1/3 t]] + Integral From 2 To Infinity [ t*f(t)dt ]

5. Jul 18, 2009

### actcs

I said that solution was

E[X] = Integral From 0 To 2 [ 2*f(t)dt ] + Integral From 2 To Infinity [ t*f(t)dt ]

Which is almost the same as you posted

It should be:

E[X] = 2 [1-Exp[-2/3]] + Integral From 2 To Infinity [ t*f(t)dt ]

Best Regards