Compute: $mathbbEleft[left(dfrac1Tint^T_0W_tdtright)^2right]$ The 2019 Stack Overflow Developer Survey Results Are InEquivalent to $beginalignint cosleft(2xright) dxendalign$?Evaluate $int frac1+cos(x)sin^2(x),operatorname d!x$Compute $int x^2 cos fracx2 mathrmdx$Evaluating: $int 3xsinleft(frac x4right) , dx$.Find $mathcalLleftt e^2tcosleft(5tright)right$For a stopping time $T$, prove that $X^T_t = mathbbEleft[X_Tmid mathcalF_tright]$Closed form for $intfracleft((x + i) betaright)^beta x^beta - 2(x^2 + 1)^beta expleft(-fracalphaA xright) , mathrmd x$Evaluate $int_0^1 fracln left(1+x+x^2+cdots +x^n right)xdx$How to find $mathbbPleft(int_0^1W_tdt>dfrac2sqrt3right)$?Find $(1-p)logleft [int^infty_0 [f(x)]^pdxright]$
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Compute: $mathbbEleft[left(dfrac1Tint^T_0W_tdtright)^2right]$
The 2019 Stack Overflow Developer Survey Results Are InEquivalent to $beginalignint cosleft(2xright) dxendalign$?Evaluate $int frac1+cos(x)sin^2(x),operatorname d!x$Compute $int x^2 cos fracx2 mathrmdx$Evaluating: $int 3xsinleft(frac x4right) , dx$.Find $mathcalLleftt e^2tcosleft(5tright)right$For a stopping time $T$, prove that $X^T_t = mathbbEleft[X_Tmid mathcalF_tright]$Closed form for $intfracleft((x + i) betaright)^beta x^beta - 2(x^2 + 1)^beta expleft(-fracalphaA xright) , mathrmd x$Evaluate $int_0^1 fracln left(1+x+x^2+cdots +x^n right)xdx$How to find $mathbbPleft(int_0^1W_tdt>dfrac2sqrt3right)$?Find $(1-p)logleft [int^infty_0 [f(x)]^pdxright]$
$begingroup$
In an interview I have been asked to solve:
$$mathbbEleft[left(dfrac1Tint^T_0W_tdtright)^2right]$$
$textbfAttempt:$
$$
beginalign
mathbbEleft[left(dfrac1Tint^T_0W_tdtright)^2right]
&= dfrac1T^2mathbbEleft[left(int^T_0W_tdtright)left(int^T_0W_uduright)right]\
&= dfrac1T^2mathbbEleft[int^T_0int^T_0W_tW_udtduright]\
&= dfrac1T^2int^T_0int^T_0mathbbEleft[W_tW_uright]dtdu \
&= dfrac1T^2int^T_0int^T_0min(u,t),dtdu
endalign
$$
Then, I couldn't continue. What would be the next steps?
integration stochastic-calculus
asked Apr 7 at 17:54
QFiQFi
615316
$endgroup$
add a comment |
$begingroup$
In an interview I have been asked to solve:
$$mathbbEleft[left(dfrac1Tint^T_0W_tdtright)^2right]$$
$textbfAttempt:$
$$
beginalign
mathbbEleft[left(dfrac1Tint^T_0W_tdtright)^2right]
&= dfrac1T^2mathbbEleft[left(int^T_0W_tdtright)left(int^T_0W_uduright)right]\
&= dfrac1T^2mathbbEleft[int^T_0int^T_0W_tW_udtduright]\
&= dfrac1T^2int^T_0int^T_0mathbbEleft[W_tW_uright]dtdu \
&= dfrac1T^2int^T_0int^T_0min(u,t),dtdu
endalign
$$
Then, I couldn't continue. What would be the next steps?
integration stochastic-calculus
asked Apr 7 at 17:54
QFiQFi
615316
$endgroup$
2
$begingroup$
Assuming the first integral is done on $u$, split the second integral as an integral on $[0,u]$ and one on $[u,T]$
$endgroup$
– dan_fulea
Apr 7 at 18:01
add a comment |
$begingroup$
In an interview I have been asked to solve:
$$mathbbEleft[left(dfrac1Tint^T_0W_tdtright)^2right]$$
$textbfAttempt:$
$$
beginalign
mathbbEleft[left(dfrac1Tint^T_0W_tdtright)^2right]
&= dfrac1T^2mathbbEleft[left(int^T_0W_tdtright)left(int^T_0W_uduright)right]\
&= dfrac1T^2mathbbEleft[int^T_0int^T_0W_tW_udtduright]\
&= dfrac1T^2int^T_0int^T_0mathbbEleft[W_tW_uright]dtdu \
&= dfrac1T^2int^T_0int^T_0min(u,t),dtdu
endalign
$$
Then, I couldn't continue. What would be the next steps?
integration stochastic-calculus
asked Apr 7 at 17:54
QFiQFi
615316
$endgroup$
In an interview I have been asked to solve:
$$mathbbEleft[left(dfrac1Tint^T_0W_tdtright)^2right]$$
$textbfAttempt:$
$$
beginalign
mathbbEleft[left(dfrac1Tint^T_0W_tdtright)^2right]
&= dfrac1T^2mathbbEleft[left(int^T_0W_tdtright)left(int^T_0W_uduright)right]\
&= dfrac1T^2mathbbEleft[int^T_0int^T_0W_tW_udtduright]\
&= dfrac1T^2int^T_0int^T_0mathbbEleft[W_tW_uright]dtdu \
&= dfrac1T^2int^T_0int^T_0min(u,t),dtdu
endalign
$$
Then, I couldn't continue. What would be the next steps?
integration stochastic-calculus
integration stochastic-calculus
asked Apr 7 at 17:54
QFiQFi
615316
asked Apr 7 at 17:54
QFiQFi
615316
asked Apr 7 at 17:54
QFiQFi
615316
asked Apr 7 at 17:54
QFiQFi
615316
asked Apr 7 at 17:54
QFiQFi
615316
615316
2
$begingroup$
Assuming the first integral is done on $u$, split the second integral as an integral on $[0,u]$ and one on $[u,T]$
$endgroup$
– dan_fulea
Apr 7 at 18:01
add a comment |
2
$begingroup$
Assuming the first integral is done on $u$, split the second integral as an integral on $[0,u]$ and one on $[u,T]$
$endgroup$
– dan_fulea
Apr 7 at 18:01
2
2
$begingroup$
Assuming the first integral is done on $u$, split the second integral as an integral on $[0,u]$ and one on $[u,T]$
$endgroup$
– dan_fulea
Apr 7 at 18:01
$begingroup$
Assuming the first integral is done on $u$, split the second integral as an integral on $[0,u]$ and one on $[u,T]$
$endgroup$
– dan_fulea
Apr 7 at 18:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Thanks to @dan_fulea hint, here are the next steps to the solution:
$$
beginalign
mathbbEleft[left(dfrac1Tint^T_0W_tdtright)^2right] &= dfrac1T^2int^T_0int^T_0min(u,t),dtdu \
&= dfrac1T^2int^T_0left(int^u_0min(u,t),dt+int^T_umin(u,t),dtright)du \
&= dfrac1T^2int^T_0left(int^u_0t,dt+int^T_uu,dtright)du\
&= dfrac1T^2int^T_0left(dfracu^22+u(T-u)right)du\
&= dfrac1T^2int^T_0left(uT-dfracu^22right)du\
&= dfrac1T^2times left(dfracT^32-dfracT^36right)\
&= dfracT3
endalign
$$
answered Apr 7 at 18:44
QFiQFi
615316
$endgroup$
add a comment |
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1 Answer
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oldest
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1 Answer
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oldest
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votes
$begingroup$
Thanks to @dan_fulea hint, here are the next steps to the solution:
$$
beginalign
mathbbEleft[left(dfrac1Tint^T_0W_tdtright)^2right] &= dfrac1T^2int^T_0int^T_0min(u,t),dtdu \
&= dfrac1T^2int^T_0left(int^u_0min(u,t),dt+int^T_umin(u,t),dtright)du \
&= dfrac1T^2int^T_0left(int^u_0t,dt+int^T_uu,dtright)du\
&= dfrac1T^2int^T_0left(dfracu^22+u(T-u)right)du\
&= dfrac1T^2int^T_0left(uT-dfracu^22right)du\
&= dfrac1T^2times left(dfracT^32-dfracT^36right)\
&= dfracT3
endalign
$$
answered Apr 7 at 18:44
QFiQFi
615316
$endgroup$
add a comment |
$begingroup$
Thanks to @dan_fulea hint, here are the next steps to the solution:
$$
beginalign
mathbbEleft[left(dfrac1Tint^T_0W_tdtright)^2right] &= dfrac1T^2int^T_0int^T_0min(u,t),dtdu \
&= dfrac1T^2int^T_0left(int^u_0min(u,t),dt+int^T_umin(u,t),dtright)du \
&= dfrac1T^2int^T_0left(int^u_0t,dt+int^T_uu,dtright)du\
&= dfrac1T^2int^T_0left(dfracu^22+u(T-u)right)du\
&= dfrac1T^2int^T_0left(uT-dfracu^22right)du\
&= dfrac1T^2times left(dfracT^32-dfracT^36right)\
&= dfracT3
endalign
$$
answered Apr 7 at 18:44
QFiQFi
615316
$endgroup$
add a comment |
$begingroup$
Thanks to @dan_fulea hint, here are the next steps to the solution:
$$
beginalign
mathbbEleft[left(dfrac1Tint^T_0W_tdtright)^2right] &= dfrac1T^2int^T_0int^T_0min(u,t),dtdu \
&= dfrac1T^2int^T_0left(int^u_0min(u,t),dt+int^T_umin(u,t),dtright)du \
&= dfrac1T^2int^T_0left(int^u_0t,dt+int^T_uu,dtright)du\
&= dfrac1T^2int^T_0left(dfracu^22+u(T-u)right)du\
&= dfrac1T^2int^T_0left(uT-dfracu^22right)du\
&= dfrac1T^2times left(dfracT^32-dfracT^36right)\
&= dfracT3
endalign
$$
answered Apr 7 at 18:44
QFiQFi
615316
$endgroup$
Thanks to @dan_fulea hint, here are the next steps to the solution:
$$
beginalign
mathbbEleft[left(dfrac1Tint^T_0W_tdtright)^2right] &= dfrac1T^2int^T_0int^T_0min(u,t),dtdu \
&= dfrac1T^2int^T_0left(int^u_0min(u,t),dt+int^T_umin(u,t),dtright)du \
&= dfrac1T^2int^T_0left(int^u_0t,dt+int^T_uu,dtright)du\
&= dfrac1T^2int^T_0left(dfracu^22+u(T-u)right)du\
&= dfrac1T^2int^T_0left(uT-dfracu^22right)du\
&= dfrac1T^2times left(dfracT^32-dfracT^36right)\
&= dfracT3
endalign
$$
answered Apr 7 at 18:44
QFiQFi
615316
answered Apr 7 at 18:44
QFiQFi
615316
answered Apr 7 at 18:44
QFiQFi
615316
answered Apr 7 at 18:44
QFiQFi
615316
615316
add a comment |
add a comment |
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$begingroup$
Assuming the first integral is done on $u$, split the second integral as an integral on $[0,u]$ and one on $[u,T]$
$endgroup$
– dan_fulea
Apr 7 at 18:01