Kirsty Gledhill

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(Indicator functions). Let ? be a sample space and let A be an event. We denote by 1_A the function defined by 1_A(?) = { 1 if ? ? A 0 otherwise }, in other words the function that is equal to one if event A occurs and that is equal to zero if event A does not occur (in which case the event A^c occurs). Show that for all ? ? ?: • 1_A(?) + 1_{A^c}(?) = 1 • 1_{A?B}(?) = 1_A(?) · 1_B(?) • 1_{A?B}(?) = 1_A(?) + 1_B(?) ? 1_{A?B}(?) • 1_{A?B}(?) = 1_A(?) + 1_B(?) provided that A and B are disjoint events. Some notation: By definition E(1_A) := P(A); that is, the expectation of the indicator function of an event is equal to the probability of that event.

          (Indicator functions). Let ? be a sample space and let A be an event. We denote by 1_A the function defined by

1_A(?) = { 1 if ? ? A
            0 otherwise },

in other words the function that is equal to one if event A occurs and that is equal to zero if event A does not occur (in which case the event A^c occurs). Show that for all ? ? ?:

• 1_A(?) + 1_{A^c}(?) = 1
• 1_{A?B}(?) = 1_A(?) · 1_B(?)
• 1_{A?B}(?) = 1_A(?) + 1_B(?) ? 1_{A?B}(?)
• 1_{A?B}(?) = 1_A(?) + 1_B(?) provided that A and B are disjoint events.

Some notation: By definition E(1_A) := P(A); that is, the expectation of the indicator function of an event is equal to the probability of that event.

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(Indicator functions). Let ? be a sample space and let A be an event. We denote by 1A the function defined by

1A(?) = 1 if ? ? A
0 otherwise ,

in other words the function that is equal to one if event A occurs and that is equal to zero if event A does not occur (in which case the event A^c occurs). Show that for all ? ? ?:

• 1A(?) + 1A^c(?) = 1
• 1A?B(?) = 1A(?) · 1B(?)
• 1A?B(?) = 1A(?) + 1B(?) ? 1A?B(?)
• 1A?B(?) = 1A(?) + 1B(?) provided that A and B are disjoint events.

Some notation: By definition E(1A) := P(A); that is, the expectation of the indicator function of an event is equal to the probability of that event.

Added by Kyle G.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Kirsty Gledhill

10/24/2022

Step 1

IA(w) + IAc(w) = 1: This is true because either the event A occurs or it does not occur. If A occurs, IA(w) = 1 and IAc(w) = 0. If A does not occur, IA(w) = 0 and IAc(w) = 1. In either case, their sum is 1. Show more…

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Transcript

00:01 Okay, so this is just some proofs about set theory.

00:03 So we've got this definition that this identity function of amiga is one if amiga is in the event a here and zero otherwise.

00:16 And this is all part of some space omega of which a is a subset of.

00:21 So the first thing it asks us to show is that 1a amiga plus 1a c, omega equals 1.

00:36 So what we're going to do here is we're going to note that this equals.

00:45 So if we're going to again do a split up omega into the same subset as here, so we're going to have omega is obviously the union of a and its complement, right? if you say this is omega and this is a, then clearly the complement of a.

01:05 Plus a gives you all of amiga.

01:09 So we're splitting our events into amiga in a and amiga not in a, or amiga an ac.

01:15 And if amiga is an a, then 1a amiga is 1, and 1 ac amiga is 0, because amiga is not an ac.

01:23 And if amiga is an ac, then 1a amiga is 0, and 1ac amiga is 1.

01:27 And we can see in both these cases it's 1, hence we've proved the statement.

01:32 Part two then asks us about when amiga is in a and b.

01:40 And so in order to answer this, we're going to consider amiga being split up into a minus b, unioned with b minus a, unioned with a intersect b, unioned with the complement of a or b.

02:02 So everything outside a and b.

02:03 And we can show that this thing.

02:05 So suppose that amiga is my big circle.

02:09 I've got b, a.

02:12 Amiga is my big square, sorry.

02:13 Then a minus b, this first bit is here.

02:20 B minus a, this second bit is this bit.

02:24 A intersect b is this bit.

02:26 And then the final one is everything else.

02:30 And so we can see that from the definition here, the definition that the question gives of for a.

02:41 So let's first consider amiga in the first bit.

02:46 So if amiga is in a but not in b, then it's not in the intersection.

02:50 So this is going to give you zero.

02:52 If amiga is in b and not a, again, it's not in the intersection, so it's going to give you zero.

02:57 If amiga is in the intersection, then that's precisely what defined.

03:02 This function is defined as giving you one in this case...

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