D**u
发帖数: 204 | 1 Heard the following interesting question.
Let x be a random variable, and y = min(x,1).
Prove that var(x) >= var(y). | z****g
发帖数: 1978 | 2 ....这个有什么好证明的，你把var的积分写出来不是显然的事么.... | D**u
发帖数: 204 | 3 can you give more detail about how to write it?
【在 z****g 的大作中提到】
: ....这个有什么好证明的，你把var的积分写出来不是显然的事么....
| B****n
发帖数: 11290 | 4 Let f(x) denote the density (or probability) of X.
The distribution of min(X,1) is the same as the distribution of the sum
P(X<=1)*G+P(X>1)*1, where the random variable G has conditional density f(x|
X<=1).
So the variance of X is equal to P(X<=1)^2*Var(G).
Var(X)>=E(Var(X|I{X<=1}))>=P(X<=1)*Var(X|X<=1), where I is an indicator
function.
So Var(X)>=P(X<=1)^2*Var(X|X<=1)=Var(min(X,1))
【在 D**u 的大作中提到】
: Heard the following interesting question.
: Let x be a random variable, and y = min(x,1).
: Prove that var(x) >= var(y).
| D**u
发帖数: 204 | 5 Based on the last line of your proof, I guess you mean that
"So the variance of min(X,1) is equal to P(X<=1)^2*Var(G)."
But I don't think this statement is correct.
For example, if X only takes values at 0 and 2,
then G only take value at a single point 0 so that Var(G) = 0.
But Var(min(X,1)) is not 0.
x|
【在 B****n 的大作中提到】
: Let f(x) denote the density (or probability) of X.
: The distribution of min(X,1) is the same as the distribution of the sum
: P(X<=1)*G+P(X>1)*1, where the random variable G has conditional density f(x|
: X<=1).
: So the variance of X is equal to P(X<=1)^2*Var(G).
: Var(X)>=E(Var(X|I{X<=1}))>=P(X<=1)*Var(X|X<=1), where I is an indicator
: function.
: So Var(X)>=P(X<=1)^2*Var(X|X<=1)=Var(min(X,1))
| x******a
发帖数: 6336 | 6 First note that Y=XI_{X<=1}+I_{X>1} and
X=Y+Z where Z=(X-1)I_{X>1}>=0, we have
EY<=1;
E(YZ)=E[(X-1)I_{X>1}]=E(Z)>=0;
it follows
Var(X)=Var(Y)+Var(Z) +2[E(YZ)-EYEZ]
>=Var(Y)+Var(Z)>=Var(Y).
【在 D**u 的大作中提到】
: Heard the following interesting question.
: Let x be a random variable, and y = min(x,1).
: Prove that var(x) >= var(y).
| D**u
发帖数: 204 | 7 很牛.
【在 x******a 的大作中提到】
: First note that Y=XI_{X<=1}+I_{X>1} and
: X=Y+Z where Z=(X-1)I_{X>1}>=0, we have
: EY<=1;
: E(YZ)=E[(X-1)I_{X>1}]=E(Z)>=0;
: it follows
: Var(X)=Var(Y)+Var(Z) +2[E(YZ)-EYEZ]
: >=Var(Y)+Var(Z)>=Var(Y).
| p*****k
发帖数: 318 | 8 note there is nothing special about 1 and min
[ as Z is simply max(x,1)-1 ], so this indicates
that a vanilla call/put is less volatile than the stock.
(basically following from xiaojiya's proof that
the min and max are positively correlated)
in BS world seems this easily follows from the fact
that delta is smaller than 1, but of course from the
proof above, the conclusion holds with no assumption
at all | D**u
发帖数: 204 | 9 This is quite an interesting interpretation.
Here is another solution:
It is not hard to check that
(x - E(x))^2 >= (y - E(x))^2 - (E(y) - E(x))^2
for any x.
Take E(.) on both sides, the left side is
var(x); and the right side is var(y). So we have
var(x) >= var(y).
【在 p*****k 的大作中提到】
: note there is nothing special about 1 and min
: [ as Z is simply max(x,1)-1 ], so this indicates
: that a vanilla call/put is less volatile than the stock.
: (basically following from xiaojiya's proof that
: the min and max are positively correlated)
: in BS world seems this easily follows from the fact
: that delta is smaller than 1, but of course from the
: proof above, the conclusion holds with no assumption
: at all
| D**u
发帖数: 204 | 10 pcasnik, can you give more detail about why delta<1 implies that
var(x) > var(y)?
Thanks.
【在 p*****k 的大作中提到】
: note there is nothing special about 1 and min
: [ as Z is simply max(x,1)-1 ], so this indicates
: that a vanilla call/put is less volatile than the stock.
: (basically following from xiaojiya's proof that
: the min and max are positively correlated)
: in BS world seems this easily follows from the fact
: that delta is smaller than 1, but of course from the
: proof above, the conclusion holds with no assumption
: at all
| t*******g
发帖数: 373 | 11 分段积分咯。
【在 D**u 的大作中提到】
: Heard the following interesting question.
: Let x be a random variable, and y = min(x,1).
: Prove that var(x) >= var(y).
| t*******g
发帖数: 373 | 12 Nice~
【在 x******a 的大作中提到】
: First note that Y=XI_{X<=1}+I_{X>1} and
: X=Y+Z where Z=(X-1)I_{X>1}>=0, we have
: EY<=1;
: E(YZ)=E[(X-1)I_{X>1}]=E(Z)>=0;
: it follows
: Var(X)=Var(Y)+Var(Z) +2[E(YZ)-EYEZ]
: >=Var(Y)+Var(Z)>=Var(Y).
| B*********s
发帖数: 306 | 13
Z is basically a call with strike price of 1, Y a covered call, and X the
underlying. So the covered call is less volatile than the underlying. This
makes sense, as you are eliminating the variation of X above 1 and put all
of the probability mass on 1 instead. This would reduce the variance.
In terms of delta in the Black-Scholes model, the underlying has a delta
of 1, and the covered call has a delta less than 1. In this sense, also,
this shows that the covered call is less risky than the underlying.
【在 x******a 的大作中提到】
: First note that Y=XI_{X<=1}+I_{X>1} and
: X=Y+Z where Z=(X-1)I_{X>1}>=0, we have
: EY<=1;
: E(YZ)=E[(X-1)I_{X>1}]=E(Z)>=0;
: it follows
: Var(X)=Var(Y)+Var(Z) +2[E(YZ)-EYEZ]
: >=Var(Y)+Var(Z)>=Var(Y).
|
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