Proving martingale property of $N_t = Z(M_twedge s - M_t wedge r)$ for martingale $M$A Stopping Theorem for Right-Continuous SubmartingalesWhy is it true that the continuous local martingale with quadratic variation “t” is a square integrable continuous martingale?Can we prove directly that $M_t$ is a martingaleMartingale property of a counting process subtracting its compensatorMartingale property of product of martingale and stochastic process.Sufficient condition for martingale propertyUniform integrability of stopped martingaleStopped local martingale as a martingaleShow that $Z(M_s wedge t - M_r wedge t)$ is a martingale when $M$ is a martingale and $Z in mathscrF_r$How can I show that the stochastic integral of a jump process w.r.t. Brownian motion is a local martingale by using this special localizing sequence?
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Proving martingale property of $N_t = Z(M_twedge s - M_t wedge r)$ for martingale $M$
A Stopping Theorem for Right-Continuous SubmartingalesWhy is it true that the continuous local martingale with quadratic variation “t” is a square integrable continuous martingale?Can we prove directly that $M_t$ is a martingaleMartingale property of a counting process subtracting its compensatorMartingale property of product of martingale and stochastic process.Sufficient condition for martingale propertyUniform integrability of stopped martingaleStopped local martingale as a martingaleShow that $Z(M_s wedge t - M_r wedge t)$ is a martingale when $M$ is a martingale and $Z in mathscrF_r$How can I show that the stochastic integral of a jump process w.r.t. Brownian motion is a local martingale by using this special localizing sequence?
$begingroup$
(Stochastic calculus and Brownian motion, LeGall, page 80).
Suppose $M = (M_t)$ is a martingale. Also, let $Z$ be a bounded random variable which is $mathcalF_r$ adapted. Then we like to show that for any $0 leq r < s$,
$$N_t = Z(M_twedge s - M_t wedge r)$$
is a martingale.
My attempt:
$Z(M_twedge s - M_t wedge r) in L_1$, should be fine since both $Z$ and $M$ are bounded. I am not sure how to show it is adapted.
Finally we need to show the martingale identity. Suppose $v geq r$, then we have
$$mathbbE Z(M_twedge s - M_t wedge r) mid mathcalF_v = Z mathbbE (M_twedge s - M_t wedge r) mid mathcalF_v = Z (M_vwedge s - M_v wedge r)$$
where in the first equality we use the fact that $Z in mathcalF_r$, and $v geq r$. This proves the result for this case. However, I don't know how to get the result for general $v$.
Thanks for you helps in advance.
probability-theory martingales local-martingales
edited Mar 2 at 18:46
saz
81.6k861127
asked Jun 8 '18 at 3:55
NickNick
293
$endgroup$
add a comment |
$begingroup$
(Stochastic calculus and Brownian motion, LeGall, page 80).
Suppose $M = (M_t)$ is a martingale. Also, let $Z$ be a bounded random variable which is $mathcalF_r$ adapted. Then we like to show that for any $0 leq r < s$,
$$N_t = Z(M_twedge s - M_t wedge r)$$
is a martingale.
My attempt:
$Z(M_twedge s - M_t wedge r) in L_1$, should be fine since both $Z$ and $M$ are bounded. I am not sure how to show it is adapted.
Finally we need to show the martingale identity. Suppose $v geq r$, then we have
$$mathbbE Z(M_twedge s - M_t wedge r) mid mathcalF_v = Z mathbbE (M_twedge s - M_t wedge r) mid mathcalF_v = Z (M_vwedge s - M_v wedge r)$$
where in the first equality we use the fact that $Z in mathcalF_r$, and $v geq r$. This proves the result for this case. However, I don't know how to get the result for general $v$.
Thanks for you helps in advance.
probability-theory martingales local-martingales
edited Mar 2 at 18:46
saz
81.6k861127
asked Jun 8 '18 at 3:55
NickNick
293
$endgroup$
$begingroup$
"and hence the bound is in $L_1$" could you elaborate why you believe that this is true? Note that $X<infty$ does not imply $X in L_1$.
$endgroup$
– saz
Jun 8 '18 at 4:20
$begingroup$
My mistake. thanks for reminding me. I edited the response. In fact, I couldn't show even the $L_1$ yet.
$endgroup$
– Nick
Jun 8 '18 at 6:57
$begingroup$
That's not surprising because it is in general wrong. If your assertion was correct, it would imply that $(M_r)_r leq t$ is a martingale for any local martingale $M$... which is clearly wrong.
$endgroup$
– saz
Jun 8 '18 at 12:53
$begingroup$
Do you impose $r leq t$ ? Or could $Z$ be rather $F_t$-adapted ? I cannot imagine how $N_t$ is adapted (with respect to $F_t$, right?) as you state it.
$endgroup$
– Leon
Jun 8 '18 at 13:10
$begingroup$
It seems too much for the general local martingale. I have modified the statement of the problem according to LeGall (page 80). The integrability should be fine in this case, but still addaptedness and martingale identity is not. Thanks again.
$endgroup$
– Nick
Jun 9 '18 at 1:56
add a comment |
$begingroup$
(Stochastic calculus and Brownian motion, LeGall, page 80).
Suppose $M = (M_t)$ is a martingale. Also, let $Z$ be a bounded random variable which is $mathcalF_r$ adapted. Then we like to show that for any $0 leq r < s$,
$$N_t = Z(M_twedge s - M_t wedge r)$$
is a martingale.
My attempt:
$Z(M_twedge s - M_t wedge r) in L_1$, should be fine since both $Z$ and $M$ are bounded. I am not sure how to show it is adapted.
Finally we need to show the martingale identity. Suppose $v geq r$, then we have
$$mathbbE Z(M_twedge s - M_t wedge r) mid mathcalF_v = Z mathbbE (M_twedge s - M_t wedge r) mid mathcalF_v = Z (M_vwedge s - M_v wedge r)$$
where in the first equality we use the fact that $Z in mathcalF_r$, and $v geq r$. This proves the result for this case. However, I don't know how to get the result for general $v$.
Thanks for you helps in advance.
probability-theory martingales local-martingales
edited Mar 2 at 18:46
saz
81.6k861127
asked Jun 8 '18 at 3:55
NickNick
293
$endgroup$
(Stochastic calculus and Brownian motion, LeGall, page 80).
Suppose $M = (M_t)$ is a martingale. Also, let $Z$ be a bounded random variable which is $mathcalF_r$ adapted. Then we like to show that for any $0 leq r < s$,
$$N_t = Z(M_twedge s - M_t wedge r)$$
is a martingale.
My attempt:
$Z(M_twedge s - M_t wedge r) in L_1$, should be fine since both $Z$ and $M$ are bounded. I am not sure how to show it is adapted.
Finally we need to show the martingale identity. Suppose $v geq r$, then we have
$$mathbbE Z(M_twedge s - M_t wedge r) mid mathcalF_v = Z mathbbE (M_twedge s - M_t wedge r) mid mathcalF_v = Z (M_vwedge s - M_v wedge r)$$
where in the first equality we use the fact that $Z in mathcalF_r$, and $v geq r$. This proves the result for this case. However, I don't know how to get the result for general $v$.
Thanks for you helps in advance.
probability-theory martingales local-martingales
probability-theory martingales local-martingales
edited Mar 2 at 18:46
saz
81.6k861127
asked Jun 8 '18 at 3:55
NickNick
293
edited Mar 2 at 18:46
saz
81.6k861127
asked Jun 8 '18 at 3:55
NickNick
293
edited Mar 2 at 18:46
saz
81.6k861127
edited Mar 2 at 18:46
saz
81.6k861127
edited Mar 2 at 18:46
saz
81.6k861127
81.6k861127
asked Jun 8 '18 at 3:55
NickNick
293
asked Jun 8 '18 at 3:55
NickNick
293
asked Jun 8 '18 at 3:55
NickNick
293
293
$begingroup$
"and hence the bound is in $L_1$" could you elaborate why you believe that this is true? Note that $X<infty$ does not imply $X in L_1$.
$endgroup$
– saz
Jun 8 '18 at 4:20
$begingroup$
My mistake. thanks for reminding me. I edited the response. In fact, I couldn't show even the $L_1$ yet.
$endgroup$
– Nick
Jun 8 '18 at 6:57
$begingroup$
That's not surprising because it is in general wrong. If your assertion was correct, it would imply that $(M_r)_r leq t$ is a martingale for any local martingale $M$... which is clearly wrong.
$endgroup$
– saz
Jun 8 '18 at 12:53
$begingroup$
Do you impose $r leq t$ ? Or could $Z$ be rather $F_t$-adapted ? I cannot imagine how $N_t$ is adapted (with respect to $F_t$, right?) as you state it.
$endgroup$
– Leon
Jun 8 '18 at 13:10
$begingroup$
It seems too much for the general local martingale. I have modified the statement of the problem according to LeGall (page 80). The integrability should be fine in this case, but still addaptedness and martingale identity is not. Thanks again.
$endgroup$
– Nick
Jun 9 '18 at 1:56
add a comment |
$begingroup$
"and hence the bound is in $L_1$" could you elaborate why you believe that this is true? Note that $X<infty$ does not imply $X in L_1$.
$endgroup$
– saz
Jun 8 '18 at 4:20
$begingroup$
My mistake. thanks for reminding me. I edited the response. In fact, I couldn't show even the $L_1$ yet.
$endgroup$
– Nick
Jun 8 '18 at 6:57
$begingroup$
That's not surprising because it is in general wrong. If your assertion was correct, it would imply that $(M_r)_r leq t$ is a martingale for any local martingale $M$... which is clearly wrong.
$endgroup$
– saz
Jun 8 '18 at 12:53
$begingroup$
Do you impose $r leq t$ ? Or could $Z$ be rather $F_t$-adapted ? I cannot imagine how $N_t$ is adapted (with respect to $F_t$, right?) as you state it.
$endgroup$
– Leon
Jun 8 '18 at 13:10
$begingroup$
It seems too much for the general local martingale. I have modified the statement of the problem according to LeGall (page 80). The integrability should be fine in this case, but still addaptedness and martingale identity is not. Thanks again.
$endgroup$
– Nick
Jun 9 '18 at 1:56
$begingroup$
"and hence the bound is in $L_1$" could you elaborate why you believe that this is true? Note that $X<infty$ does not imply $X in L_1$.
$endgroup$
– saz
Jun 8 '18 at 4:20
$begingroup$
"and hence the bound is in $L_1$" could you elaborate why you believe that this is true? Note that $X<infty$ does not imply $X in L_1$.
$endgroup$
– saz
Jun 8 '18 at 4:20
$begingroup$
My mistake. thanks for reminding me. I edited the response. In fact, I couldn't show even the $L_1$ yet.
$endgroup$
– Nick
Jun 8 '18 at 6:57
$begingroup$
My mistake. thanks for reminding me. I edited the response. In fact, I couldn't show even the $L_1$ yet.
$endgroup$
– Nick
Jun 8 '18 at 6:57
$begingroup$
That's not surprising because it is in general wrong. If your assertion was correct, it would imply that $(M_r)_r leq t$ is a martingale for any local martingale $M$... which is clearly wrong.
$endgroup$
– saz
Jun 8 '18 at 12:53
$begingroup$
That's not surprising because it is in general wrong. If your assertion was correct, it would imply that $(M_r)_r leq t$ is a martingale for any local martingale $M$... which is clearly wrong.
$endgroup$
– saz
Jun 8 '18 at 12:53
$begingroup$
Do you impose $r leq t$ ? Or could $Z$ be rather $F_t$-adapted ? I cannot imagine how $N_t$ is adapted (with respect to $F_t$, right?) as you state it.
$endgroup$
– Leon
Jun 8 '18 at 13:10
$begingroup$
Do you impose $r leq t$ ? Or could $Z$ be rather $F_t$-adapted ? I cannot imagine how $N_t$ is adapted (with respect to $F_t$, right?) as you state it.
$endgroup$
– Leon
Jun 8 '18 at 13:10
$begingroup$
It seems too much for the general local martingale. I have modified the statement of the problem according to LeGall (page 80). The integrability should be fine in this case, but still addaptedness and martingale identity is not. Thanks again.
$endgroup$
– Nick
Jun 9 '18 at 1:56
$begingroup$
It seems too much for the general local martingale. I have modified the statement of the problem according to LeGall (page 80). The integrability should be fine in this case, but still addaptedness and martingale identity is not. Thanks again.
$endgroup$
– Nick
Jun 9 '18 at 1:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First of all, we note that
$$N_t = begincases 0, & t leq r, \ Z (M_t-M_r) & t in (r,s), \ Z (M_s-M_r), & t geq s. endcases tag1$$
Since $Z$ is, by assumption, $mathcalF_r$-measurable and $(M_t)_t geq 0$ is $(mathcalF_t)_t geq 0$-adapted, this implies, in particular, that $(N_t)_t geq 0$ is $(mathcalF_t)_t geq 0$-adapted. Moreover, $M_t in L^1$ together with the boundedness of $Z$ entails $N_t in L^1$ for all $t geq 0$. It remains to check that
$$mathbbE(N_t mid mathcalF_u) = N_u quad textfor all $u leq t$. tag2$$
We consider two cases separately. If $t geq u geq r$ then it follows from the $mathcalF_r$-measurability of $Z$ that
$$mathbbE(N_t mid mathcalF_u) = Z mathbbE(M_t wedge s-M_t wedge r mid mathcalF_u).$$
Because of the martingal property of $(M_t)_t geq 0$ we have
$$ mathbbE(M_t wedge s-M_t wedge r mid mathcalF_u) = M_t wedge s wedge u - M_t wedge r wedge u stackrelt geq u= M_u wedge s - M_u wedge r,$$
and so
$$mathbbE(N_t mid mathcalF_u) = N_u, qquad r leq u leq t. $$
In particular,
$$mathbbE(N_t mid mathcalF_r) = N_r stackrel(1)= 0. tag3$$
It remains to consider the case $u < r$. If $t leq r$ then, by $(1)$, $N_t = 0$ and so
$$mathbbE(N_t mid mathcalF_u)=0=N_u.$$
Suppose now that $t geq r$. Since the tower property gives
$$mathbbE(N_t mid mathcalF_u) = mathbbE bigg[ mathbbE(N_t mid mathcalF_r) mid mathcalF_u bigg], $$
we conclude from $(3)$ that
$$mathbbE(N_t mid mathcalF_u) = 0 = N_u.$$
This finishes the proof of $(2)$.
answered Mar 2 at 18:43
sazsaz
81.6k861127
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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votes
$begingroup$
First of all, we note that
$$N_t = begincases 0, & t leq r, \ Z (M_t-M_r) & t in (r,s), \ Z (M_s-M_r), & t geq s. endcases tag1$$
Since $Z$ is, by assumption, $mathcalF_r$-measurable and $(M_t)_t geq 0$ is $(mathcalF_t)_t geq 0$-adapted, this implies, in particular, that $(N_t)_t geq 0$ is $(mathcalF_t)_t geq 0$-adapted. Moreover, $M_t in L^1$ together with the boundedness of $Z$ entails $N_t in L^1$ for all $t geq 0$. It remains to check that
$$mathbbE(N_t mid mathcalF_u) = N_u quad textfor all $u leq t$. tag2$$
We consider two cases separately. If $t geq u geq r$ then it follows from the $mathcalF_r$-measurability of $Z$ that
$$mathbbE(N_t mid mathcalF_u) = Z mathbbE(M_t wedge s-M_t wedge r mid mathcalF_u).$$
Because of the martingal property of $(M_t)_t geq 0$ we have
$$ mathbbE(M_t wedge s-M_t wedge r mid mathcalF_u) = M_t wedge s wedge u - M_t wedge r wedge u stackrelt geq u= M_u wedge s - M_u wedge r,$$
and so
$$mathbbE(N_t mid mathcalF_u) = N_u, qquad r leq u leq t. $$
In particular,
$$mathbbE(N_t mid mathcalF_r) = N_r stackrel(1)= 0. tag3$$
It remains to consider the case $u < r$. If $t leq r$ then, by $(1)$, $N_t = 0$ and so
$$mathbbE(N_t mid mathcalF_u)=0=N_u.$$
Suppose now that $t geq r$. Since the tower property gives
$$mathbbE(N_t mid mathcalF_u) = mathbbE bigg[ mathbbE(N_t mid mathcalF_r) mid mathcalF_u bigg], $$
we conclude from $(3)$ that
$$mathbbE(N_t mid mathcalF_u) = 0 = N_u.$$
This finishes the proof of $(2)$.
answered Mar 2 at 18:43
sazsaz
81.6k861127
$endgroup$
add a comment |
$begingroup$
First of all, we note that
$$N_t = begincases 0, & t leq r, \ Z (M_t-M_r) & t in (r,s), \ Z (M_s-M_r), & t geq s. endcases tag1$$
Since $Z$ is, by assumption, $mathcalF_r$-measurable and $(M_t)_t geq 0$ is $(mathcalF_t)_t geq 0$-adapted, this implies, in particular, that $(N_t)_t geq 0$ is $(mathcalF_t)_t geq 0$-adapted. Moreover, $M_t in L^1$ together with the boundedness of $Z$ entails $N_t in L^1$ for all $t geq 0$. It remains to check that
$$mathbbE(N_t mid mathcalF_u) = N_u quad textfor all $u leq t$. tag2$$
We consider two cases separately. If $t geq u geq r$ then it follows from the $mathcalF_r$-measurability of $Z$ that
$$mathbbE(N_t mid mathcalF_u) = Z mathbbE(M_t wedge s-M_t wedge r mid mathcalF_u).$$
Because of the martingal property of $(M_t)_t geq 0$ we have
$$ mathbbE(M_t wedge s-M_t wedge r mid mathcalF_u) = M_t wedge s wedge u - M_t wedge r wedge u stackrelt geq u= M_u wedge s - M_u wedge r,$$
and so
$$mathbbE(N_t mid mathcalF_u) = N_u, qquad r leq u leq t. $$
In particular,
$$mathbbE(N_t mid mathcalF_r) = N_r stackrel(1)= 0. tag3$$
It remains to consider the case $u < r$. If $t leq r$ then, by $(1)$, $N_t = 0$ and so
$$mathbbE(N_t mid mathcalF_u)=0=N_u.$$
Suppose now that $t geq r$. Since the tower property gives
$$mathbbE(N_t mid mathcalF_u) = mathbbE bigg[ mathbbE(N_t mid mathcalF_r) mid mathcalF_u bigg], $$
we conclude from $(3)$ that
$$mathbbE(N_t mid mathcalF_u) = 0 = N_u.$$
This finishes the proof of $(2)$.
answered Mar 2 at 18:43
sazsaz
81.6k861127
$endgroup$
add a comment |
$begingroup$
First of all, we note that
$$N_t = begincases 0, & t leq r, \ Z (M_t-M_r) & t in (r,s), \ Z (M_s-M_r), & t geq s. endcases tag1$$
Since $Z$ is, by assumption, $mathcalF_r$-measurable and $(M_t)_t geq 0$ is $(mathcalF_t)_t geq 0$-adapted, this implies, in particular, that $(N_t)_t geq 0$ is $(mathcalF_t)_t geq 0$-adapted. Moreover, $M_t in L^1$ together with the boundedness of $Z$ entails $N_t in L^1$ for all $t geq 0$. It remains to check that
$$mathbbE(N_t mid mathcalF_u) = N_u quad textfor all $u leq t$. tag2$$
We consider two cases separately. If $t geq u geq r$ then it follows from the $mathcalF_r$-measurability of $Z$ that
$$mathbbE(N_t mid mathcalF_u) = Z mathbbE(M_t wedge s-M_t wedge r mid mathcalF_u).$$
Because of the martingal property of $(M_t)_t geq 0$ we have
$$ mathbbE(M_t wedge s-M_t wedge r mid mathcalF_u) = M_t wedge s wedge u - M_t wedge r wedge u stackrelt geq u= M_u wedge s - M_u wedge r,$$
and so
$$mathbbE(N_t mid mathcalF_u) = N_u, qquad r leq u leq t. $$
In particular,
$$mathbbE(N_t mid mathcalF_r) = N_r stackrel(1)= 0. tag3$$
It remains to consider the case $u < r$. If $t leq r$ then, by $(1)$, $N_t = 0$ and so
$$mathbbE(N_t mid mathcalF_u)=0=N_u.$$
Suppose now that $t geq r$. Since the tower property gives
$$mathbbE(N_t mid mathcalF_u) = mathbbE bigg[ mathbbE(N_t mid mathcalF_r) mid mathcalF_u bigg], $$
we conclude from $(3)$ that
$$mathbbE(N_t mid mathcalF_u) = 0 = N_u.$$
This finishes the proof of $(2)$.
answered Mar 2 at 18:43
sazsaz
81.6k861127
$endgroup$
First of all, we note that
$$N_t = begincases 0, & t leq r, \ Z (M_t-M_r) & t in (r,s), \ Z (M_s-M_r), & t geq s. endcases tag1$$
Since $Z$ is, by assumption, $mathcalF_r$-measurable and $(M_t)_t geq 0$ is $(mathcalF_t)_t geq 0$-adapted, this implies, in particular, that $(N_t)_t geq 0$ is $(mathcalF_t)_t geq 0$-adapted. Moreover, $M_t in L^1$ together with the boundedness of $Z$ entails $N_t in L^1$ for all $t geq 0$. It remains to check that
$$mathbbE(N_t mid mathcalF_u) = N_u quad textfor all $u leq t$. tag2$$
We consider two cases separately. If $t geq u geq r$ then it follows from the $mathcalF_r$-measurability of $Z$ that
$$mathbbE(N_t mid mathcalF_u) = Z mathbbE(M_t wedge s-M_t wedge r mid mathcalF_u).$$
Because of the martingal property of $(M_t)_t geq 0$ we have
$$ mathbbE(M_t wedge s-M_t wedge r mid mathcalF_u) = M_t wedge s wedge u - M_t wedge r wedge u stackrelt geq u= M_u wedge s - M_u wedge r,$$
and so
$$mathbbE(N_t mid mathcalF_u) = N_u, qquad r leq u leq t. $$
In particular,
$$mathbbE(N_t mid mathcalF_r) = N_r stackrel(1)= 0. tag3$$
It remains to consider the case $u < r$. If $t leq r$ then, by $(1)$, $N_t = 0$ and so
$$mathbbE(N_t mid mathcalF_u)=0=N_u.$$
Suppose now that $t geq r$. Since the tower property gives
$$mathbbE(N_t mid mathcalF_u) = mathbbE bigg[ mathbbE(N_t mid mathcalF_r) mid mathcalF_u bigg], $$
we conclude from $(3)$ that
$$mathbbE(N_t mid mathcalF_u) = 0 = N_u.$$
This finishes the proof of $(2)$.
answered Mar 2 at 18:43
sazsaz
81.6k861127
edited yesterday
answered Mar 2 at 18:43
sazsaz
81.6k861127
answered Mar 2 at 18:43
sazsaz
81.6k861127
answered Mar 2 at 18:43
sazsaz
81.6k861127
81.6k861127
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$begingroup$
"and hence the bound is in $L_1$" could you elaborate why you believe that this is true? Note that $X<infty$ does not imply $X in L_1$.
$endgroup$
– saz
Jun 8 '18 at 4:20
$begingroup$
My mistake. thanks for reminding me. I edited the response. In fact, I couldn't show even the $L_1$ yet.
$endgroup$
– Nick
Jun 8 '18 at 6:57
$begingroup$
That's not surprising because it is in general wrong. If your assertion was correct, it would imply that $(M_r)_r leq t$ is a martingale for any local martingale $M$... which is clearly wrong.
$endgroup$
– saz
Jun 8 '18 at 12:53
$begingroup$
Do you impose $r leq t$ ? Or could $Z$ be rather $F_t$-adapted ? I cannot imagine how $N_t$ is adapted (with respect to $F_t$, right?) as you state it.
$endgroup$
– Leon
Jun 8 '18 at 13:10
$begingroup$
It seems too much for the general local martingale. I have modified the statement of the problem according to LeGall (page 80). The integrability should be fine in this case, but still addaptedness and martingale identity is not. Thanks again.
$endgroup$
– Nick
Jun 9 '18 at 1:56