Question

Hi I'm working on a probability problem that involves 3 parts. I know how to do part a and b, but I'm stuck on part c. I was wondering if I can get some help. I'll post the question and the attempt for the solution of part c at the bottom. Any help will be great. Thanks! Will give medal.

Answer 1

Let X=Y-Z where Y and Z are nonnegative random variables such that YZ=0.
(a) show that\[cov(Y,Z)\le0\]
(b)show that \[var(X)\ge var(X)+var(Z) \]
(c) Using the result of part (b) show that \[var(X)\ge var(\max\{0,X\})+var(\max\{0,-X\})\]

Answer 2

We never learned variance of the maximum of two random variables in my class. After reading online I found something that stated \[var(\max X_i)\le \sum_{i} var(X_i)\] for iid variables. However I couldn't find any other formula.

So my attempt for part (c) was to show that \[var(\max\{0,X\})+var(\max\{0,-X\})\le var(0)+var(X)+var(-X)+var(0)\]\[=var(X)+var(-X)=var(X)+var(X)= 2var(X)\]
However this goes against what I'm trying to show. I can't seem to find another formula to implement that will allow me to show that. Am I on the right track or far off? Thank you!

Answer 3

does YZ=0. mean when Y is >0 then Z=0 and vice versa?

also, is this true
\[var(X)\ge var(X)+var(Z) \]
aren't variances always positive?

Answer 4

Yes one of them can be zero or both because they are nonnegative

Sorry I made a mistake it was supposed to be var(x)>=Var(y)+var(z)

Answer 5

I'm not sure if they are always positive. I'll try to check my book

Answer 6

var = std^2

Answer 7

so 0 or positive

Answer 8

I see. However, I'm not sure how the variance always being positive helps?

Answer 9

X is non-negative, so \( 0 \le X\)
the max of 0 and X should be X ?
i.e. max(0,X)= X
so var(max(0,X))= var(X)

Answer 10

So for var(max(0,-x)) it would be var(0)?
Would that give the it is less than or equal to var(x)+var(0) which is var(x)?

Answer 11

Since x is nonnegative 0 is greater than -x?

Answer 12

But if Z is greater than y wouldn't x be negative?

Answer 13

var(0) is 0 (i.e. the variance of a string of constants)

Answer 14

var(x)+var(0)= var(x)
However is there a possibility that var(max(0,-x))=var(-x)?

Answer 15

max(0,-x)
returns the max of 0 or -x
x is 0 or +, so -x is 0 or -
if x is 0 the max of (0,0) is 0
if x is neg, the max of (0, neg) is 0
so you always get 0

Answer 16

are sure you don't have a typo in the part c?
it would make more sense if they reference Y or Z (and not X) on the right-hand side.

Answer 17

Why wouldn't it be -(-value of x) meaning -x is the max?

No typo. That also through me off because I don't see how to incorporate them.

Answer 18

I should have said:
-x is 0 or neg
max(0,-x) will always be max(0,0) or max(0, neg number)
in either case you get 0

Answer 19

May I ask how you know that x should be nonnegative? Since the problem states that y and z are nonnegative how does that imply x is also nonnegative?

Answer 20

oh, I misread that as all were non-negative.

Answer 21

Does that change the problem?

Answer 22

Do I have to do more analysis on Y and Z?

Answer 23

yes, it looks like we should (at least it makes more sense now)

Answer 24

So if both y and z are 0 than the max is var(0) for both?
If Y greater than Z then Z must be 0 and X is nonnegative meaning Var(x)+var(0)=var(x)
and if z is greater that means var(0)+var(-x)=var(x)?

Answer 25

meaning it can equal 0 or var(x)

Answer 26

so var(x) is greater than or equal to? Or is my analysis wrong?

Answer 27

I have to think about it, but it seems
max(0,x) means max(0,y)
because y-z is y or -z. y is + , -z is neg,

Answer 28

and max(0,y) is y

Answer 29

That makes sense to me.
But when I try to do max(0,-x) using your analysis I'm not sure how it works?
So y would probably be 0 ands would be -z meaning -x =z?

Answer 30

exactly

Answer 31

Since y is nonnegative. I understand that one.

Answer 32

So is this all the analysis that would be required?

Answer 33

Also max(0,z)=z for the same reason

Answer 34

yes
so we get from
v(x) >= v( max(0,x) ) + v( max(0,-x) )
v(x) >= v( max(0,y) ) + v( max(0,z) )
v(x) >= v(y) + v(z)
which is proved in part b.

Answer 35

Thank you so much! I understand it now. I would have never though of how to implement y or z though. Thank you once again:D

Answer 36

it took a while, but muddling along sometimes works.

Answer 37

Yes. I think I would have kept trying what I was doing over and over again and I would have gotten nowhere though

Answer 38

Thank you and have a nice day!

Answer 39

yw