l****d
发帖数: 55 | 1 10. x martingale? sin(x) martingale or not?
it has a drift term -sin(x) when Ito lemma is applied. So it's not a
martingale. Is it right?
Having seen couple of similar questions,I wonder if someone can give an
example that when X is martingale, f(x) is also a martingale. I know X^2-n
is
but X^3 is not according to the the green book. My understanding of
martingale
is quite superficial at this point.
18.x,y~N(0,1), independent, what is E(x|x+y=1), what about variance
E(X)+E(Y)=E(X)+E(1-Y)=1=2E(X)====>E(X)=1/2, is it right?
What about variance? Can somebody give a shot at it? It seems to hard to
calculate E(X^2). | j********t
发帖数: 97 | 2 10. I think a process with zero drift term is just necessary condition of
martingale, but not sufficient condition. For example,
X(t) = cosh(aW(t))exp(-a^2 t/2) has zero drift in SDE notation but isn't a
martingale according to martingale definition.
18.
E(X^2 | X+Y=1)
= E(X^2, X+Y=1) / E(X^2) By conditional expectation definition,
= E(X^2,Y=1-X) / E(X^2)
= E(X^2)E(1-X) / E(X^2) since X,Y are indepedent
= 1- E(X) = 0.5
Var(X | X+Y=1) = 0.5 - 0.5^2 = 0.25 | k*******d
发帖数: 1340 | 3 E(X^2 | X+Y=1)
= E(X^2, X+Y=1) / E(X^2) By conditional expectation definition
这个是怎么得出来的？ E[X^2,X+Y=1]的数学定义是什么？我不是很明白啊
我觉得这道题目最直接的做法就是，定义U=X+Y, [U,X]是[X,Y]的linear
transformation，可以求出[U,X]的joint distribution, E[U]=E[X]=0, \rho = 1/
sqrt(2), sigma_U=sqrt(2)
然后condition on U,conditional distribution of X还是个normal,我算出来E[X|U=1
]是1/2， variance也是1/2
这是稳妥的做法，当然可能有更快捷的做法 | x******a
发帖数: 6336 | 4 FOR 18, noting that x-y is independent of x+y, so we have
2=E(x-y)^2=E((x-y)^2|x+y=1)=E(x^2|x+y=1)+ E(y^2|x+y=1)-2E(xy|x+y=1)
and 1=E((x+y)^2|x+y=1)=E(x^2|x+y=1)+E(y^2|x+y=1)+2E(xy|x+y=1)
sum up and by symmetry,
3=4E(x^2|x+y=1)
E(x^2|x+y=1)=3/4.
anything wrong?
【在 l****d 的大作中提到】
: 10. x martingale? sin(x) martingale or not?
: it has a drift term -sin(x) when Ito lemma is applied. So it's not a
: martingale. Is it right?
: Having seen couple of similar questions,I wonder if someone can give an
: example that when X is martingale, f(x) is also a martingale. I know X^2-n
: is
: but X^3 is not according to the the green book. My understanding of
: martingale
: is quite superficial at this point.
: 18.x,y~N(0,1), independent, what is E(x|x+y=1), what about variance
| s****p
发帖数: 19 | 5 Correct, but can be still simpler. Because x=[(x-y)+(x+y)]/2 and x-y and x+y
are independent, we directly have
Var(x|x+y=1)=Var((x-y)/2)=1/2
【在 x******a 的大作中提到】
: FOR 18, noting that x-y is independent of x+y, so we have
: 2=E(x-y)^2=E((x-y)^2|x+y=1)=E(x^2|x+y=1)+ E(y^2|x+y=1)-2E(xy|x+y=1)
: and 1=E((x+y)^2|x+y=1)=E(x^2|x+y=1)+E(y^2|x+y=1)+2E(xy|x+y=1)
: sum up and by symmetry,
: 3=4E(x^2|x+y=1)
: E(x^2|x+y=1)=3/4.
: anything wrong?
| k*******d
发帖数: 1340 | 6 You are right and this is smarter than me.
x+y
【在 s****p 的大作中提到】
: Correct, but can be still simpler. Because x=[(x-y)+(x+y)]/2 and x-y and x+y
: are independent, we directly have
: Var(x|x+y=1)=Var((x-y)/2)=1/2
| z****u
发帖数: 185 | 7 I must be stupid, but can you elaborate why X(t) is not a martingale
according
to its definition?
【在 j********t 的大作中提到】
: 10. I think a process with zero drift term is just necessary condition of
: martingale, but not sufficient condition. For example,
: X(t) = cosh(aW(t))exp(-a^2 t/2) has zero drift in SDE notation but isn't a
: martingale according to martingale definition.
: 18.
: E(X^2 | X+Y=1)
: = E(X^2, X+Y=1) / E(X^2) By conditional expectation definition,
: = E(X^2,Y=1-X) / E(X^2)
: = E(X^2)E(1-X) / E(X^2) since X,Y are indepedent
: = 1- E(X) = 0.5
| x******a
发帖数: 6336 | 8 I am not sure if it is cosh(aW(t)) or cos(aW(t))?
【在 z****u 的大作中提到】
: I must be stupid, but can you elaborate why X(t) is not a martingale
: according
: to its definition?
| j********t
发帖数: 97 | 9 Your solution looks simple. But how to prove x-y and x+y are independent?
+y
【在 s****p 的大作中提到】
: Correct, but can be still simpler. Because x=[(x-y)+(x+y)]/2 and x-y and x+y
: are independent, we directly have
: Var(x|x+y=1)=Var((x-y)/2)=1/2
| t*******y
发帖数: 637 | 10 uncorrelated and joint normal吧
【在 j********t 的大作中提到】
: Your solution looks simple. But how to prove x-y and x+y are independent?
:
: +y
| j********t
发帖数: 97 | 11 In general, uncorrelated doesn't mean independent. Is this question a
special case? Can you show more
details? Thanks.
【在 t*******y 的大作中提到】
: uncorrelated and joint normal吧
| t*******y
发帖数: 637 | 12 uncorrelated + joint normal means independet
【在 j********t 的大作中提到】
: In general, uncorrelated doesn't mean independent. Is this question a
: special case? Can you show more
: details? Thanks.
|
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