characteristic function for independent $X$ and $Y$
Clash Royale CLAN TAG#URR8PPP
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$X$ and $Y$ are independent, identically distributed random variables with mean $0$, variance $1$ and characteristic function $phi$, If $X+Y$ and $X-Y$ are independent, prove that $$phi(2t)=phi(t)^3phi(-t).$$By making the substitution $gamma(t)=phi(t)/phi(-t)$ or otherwise, show that, for any positive integer $n$,$$phi(t)=left1-frac 12left(fract2^nright)^2+oleft(left[frac t2^nright]^2right)right^4^n$$ Hence, find the common distribution of $X$ and $Y$.
I know how to do the first part. However for the second part, I have no idea at all. Can anyone teach me? Thanks.
probability-theory characteristic-functions
asked Aug 15 at 8:57
Kenneth Nye
16210
add a comment |Â
up vote
4
down vote
favorite
$X$ and $Y$ are independent, identically distributed random variables with mean $0$, variance $1$ and characteristic function $phi$, If $X+Y$ and $X-Y$ are independent, prove that $$phi(2t)=phi(t)^3phi(-t).$$By making the substitution $gamma(t)=phi(t)/phi(-t)$ or otherwise, show that, for any positive integer $n$,$$phi(t)=left1-frac 12left(fract2^nright)^2+oleft(left[frac t2^nright]^2right)right^4^n$$ Hence, find the common distribution of $X$ and $Y$.
I know how to do the first part. However for the second part, I have no idea at all. Can anyone teach me? Thanks.
probability-theory characteristic-functions
asked Aug 15 at 8:57
Kenneth Nye
16210
1
Not sure yet how to solve it but you can start from the observation that for every positive integer $n$, $gamma(2^n t) = gamma(t)^2n$
– Ant
Aug 15 at 9:08
@Ant Just realize how you get that observation and remember it appeared somewhere in the textbook. I will see how...
– Kenneth Nye
Aug 15 at 9:19
For the case of a symmetric distribution this is fairly easy. Expanding the characteristic function around $0$ and taking the limit shows that the common distribution is normal. How one can use $gamma $ to handle non-symmetric case beats me.
– Kavi Rama Murthy
Aug 15 at 9:31
@Ant I figured out the answer. Thanks.
– Kenneth Nye
Aug 15 at 10:21
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
$X$ and $Y$ are independent, identically distributed random variables with mean $0$, variance $1$ and characteristic function $phi$, If $X+Y$ and $X-Y$ are independent, prove that $$phi(2t)=phi(t)^3phi(-t).$$By making the substitution $gamma(t)=phi(t)/phi(-t)$ or otherwise, show that, for any positive integer $n$,$$phi(t)=left1-frac 12left(fract2^nright)^2+oleft(left[frac t2^nright]^2right)right^4^n$$ Hence, find the common distribution of $X$ and $Y$.
I know how to do the first part. However for the second part, I have no idea at all. Can anyone teach me? Thanks.
probability-theory characteristic-functions
asked Aug 15 at 8:57
Kenneth Nye
16210
$X$ and $Y$ are independent, identically distributed random variables with mean $0$, variance $1$ and characteristic function $phi$, If $X+Y$ and $X-Y$ are independent, prove that $$phi(2t)=phi(t)^3phi(-t).$$By making the substitution $gamma(t)=phi(t)/phi(-t)$ or otherwise, show that, for any positive integer $n$,$$phi(t)=left1-frac 12left(fract2^nright)^2+oleft(left[frac t2^nright]^2right)right^4^n$$ Hence, find the common distribution of $X$ and $Y$.
I know how to do the first part. However for the second part, I have no idea at all. Can anyone teach me? Thanks.
probability-theory characteristic-functions
asked Aug 15 at 8:57
Kenneth Nye
16210
asked Aug 15 at 8:57
Kenneth Nye
16210
asked Aug 15 at 8:57
Kenneth Nye
16210
asked Aug 15 at 8:57
Kenneth Nye
16210
16210
1
Not sure yet how to solve it but you can start from the observation that for every positive integer $n$, $gamma(2^n t) = gamma(t)^2n$
– Ant
Aug 15 at 9:08
@Ant Just realize how you get that observation and remember it appeared somewhere in the textbook. I will see how...
– Kenneth Nye
Aug 15 at 9:19
For the case of a symmetric distribution this is fairly easy. Expanding the characteristic function around $0$ and taking the limit shows that the common distribution is normal. How one can use $gamma $ to handle non-symmetric case beats me.
– Kavi Rama Murthy
Aug 15 at 9:31
@Ant I figured out the answer. Thanks.
– Kenneth Nye
Aug 15 at 10:21
add a comment |Â
1
Not sure yet how to solve it but you can start from the observation that for every positive integer $n$, $gamma(2^n t) = gamma(t)^2n$
– Ant
Aug 15 at 9:08
@Ant Just realize how you get that observation and remember it appeared somewhere in the textbook. I will see how...
– Kenneth Nye
Aug 15 at 9:19
For the case of a symmetric distribution this is fairly easy. Expanding the characteristic function around $0$ and taking the limit shows that the common distribution is normal. How one can use $gamma $ to handle non-symmetric case beats me.
– Kavi Rama Murthy
Aug 15 at 9:31
@Ant I figured out the answer. Thanks.
– Kenneth Nye
Aug 15 at 10:21
1
1
Not sure yet how to solve it but you can start from the observation that for every positive integer $n$, $gamma(2^n t) = gamma(t)^2n$
– Ant
Aug 15 at 9:08
Not sure yet how to solve it but you can start from the observation that for every positive integer $n$, $gamma(2^n t) = gamma(t)^2n$
– Ant
Aug 15 at 9:08
@Ant Just realize how you get that observation and remember it appeared somewhere in the textbook. I will see how...
– Kenneth Nye
Aug 15 at 9:19
@Ant Just realize how you get that observation and remember it appeared somewhere in the textbook. I will see how...
– Kenneth Nye
Aug 15 at 9:19
For the case of a symmetric distribution this is fairly easy. Expanding the characteristic function around $0$ and taking the limit shows that the common distribution is normal. How one can use $gamma $ to handle non-symmetric case beats me.
– Kavi Rama Murthy
Aug 15 at 9:31
For the case of a symmetric distribution this is fairly easy. Expanding the characteristic function around $0$ and taking the limit shows that the common distribution is normal. How one can use $gamma $ to handle non-symmetric case beats me.
– Kavi Rama Murthy
Aug 15 at 9:31
@Ant I figured out the answer. Thanks.
– Kenneth Nye
Aug 15 at 10:21
@Ant I figured out the answer. Thanks.
– Kenneth Nye
Aug 15 at 10:21
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
From independence we have $varphi_X+Y+(X-Y)=varphi_2X$ and thus
$$
varphi(2t) = varphi(t)^2(varphi(t)varphi(-t)) = varphi(t)^3varphi(-t).
$$
Set $gamma(t) = varphi(t)/varphi(-t)$, then
$$
gamma(2t) = fracvarphi(2t)varphi(-2t) = fracvarphi(t)^3varphi(-t)varphi(-t)^3varphi(t) =gamma(t)^2.
$$
It follows that $gamma(t)=gamma(t/2)^2$ and by induction, $gamma(t) = gamma(t/2^n)^2^n$ for nonnegative integers $n$. Moreover, $varphi'(0)=imathbb E[X]=0$ and $varphi''(0)=i^2mathbb E[X^2] = -1$, so as $hto0$ we have by Taylor expansions
$$
gamma(h) = fracvarphi(h)varphi(-h) = frac1-frac12h^2+o(h^2)1-frac12h^2+o(h^2) = 1 + o(h)^2.
$$
It follows that for large $n$, $$gamma(t) = (1+o(t^2/2^2n)^2^nstackrelntoinftylongrightarrow1,$$
and so $gammaequiv1$, yielding $varphi(t)=varphi(-t)$. Therefore $varphi(2t)=varphi(t)^4$, hence $varphi(t)=varphi(t/2)^4$ and by induction, $$varphi(t)=varphi(t/2^n)^4^n,ngeqslant 1.$$
Using Taylor expansions we see that for large $n$,
$$
varphi(t) = left(1-frac12left(frac t2^nright)^2+oleft(left(frac t2^nright)^2right)right)^4^nstackrelntoinftylongrightarrow expleft(-frac12 t^2right)
$$
and hence $X$ and $Y$ have standard normal distribution.
answered Aug 15 at 10:21
Math1000
18.5k31644
1
For the part to show that $gamma(t)=1+o(t^2)$ you can actually differentiate it once and plug in zero to show that it does not have t term.
– Kenneth Nye
Aug 15 at 10:27
add a comment |Â
up vote
2
down vote
The substitution $gamma(t)$ has the following properties:$$gamma(t) = phi(t)/phi(-t)=fracphi^3(t/2)phi(-t/2)phi(t/2)phi^3(-t/2)=gamma^2(t/2)$$ therefore,$$gamma(t)=lim_ntoinftygamma^2^n(t/2^n)$$
It is easy to show that $$gamma(0)=1$$
$$gamma'(0)=0$$ That means the maclaurin expansion as $tto0$ will be $$gamma(t)=1+o(t^2) $$ thus $$gamma(t)=lim_ntoinftyleft[1+o(t^2/4^n)right]^2^n=1$$ Hence, $$phi(t)=phi(-t)$$ And further that $$phi(2t)=phi^4(t)$$ Therefore, $$phi(t)=phi^4^n(t/2^n)$$ We know that $$phi(t)=1+E(X)ti-fracE(X^2)2t^2+o(t^2)=1-fract^22+o(t^2)$$ then we get that long expression. By taking limit on it we get $$phi(t)=lim_ntoinftyleft1-frac12fract^24^nright^4^n=e^-frac12t^2$$
answered Aug 15 at 10:21
Kenneth Nye
16210
Very nice work! :-)
– Ant
Aug 15 at 10:23
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
From independence we have $varphi_X+Y+(X-Y)=varphi_2X$ and thus
$$
varphi(2t) = varphi(t)^2(varphi(t)varphi(-t)) = varphi(t)^3varphi(-t).
$$
Set $gamma(t) = varphi(t)/varphi(-t)$, then
$$
gamma(2t) = fracvarphi(2t)varphi(-2t) = fracvarphi(t)^3varphi(-t)varphi(-t)^3varphi(t) =gamma(t)^2.
$$
It follows that $gamma(t)=gamma(t/2)^2$ and by induction, $gamma(t) = gamma(t/2^n)^2^n$ for nonnegative integers $n$. Moreover, $varphi'(0)=imathbb E[X]=0$ and $varphi''(0)=i^2mathbb E[X^2] = -1$, so as $hto0$ we have by Taylor expansions
$$
gamma(h) = fracvarphi(h)varphi(-h) = frac1-frac12h^2+o(h^2)1-frac12h^2+o(h^2) = 1 + o(h)^2.
$$
It follows that for large $n$, $$gamma(t) = (1+o(t^2/2^2n)^2^nstackrelntoinftylongrightarrow1,$$
and so $gammaequiv1$, yielding $varphi(t)=varphi(-t)$. Therefore $varphi(2t)=varphi(t)^4$, hence $varphi(t)=varphi(t/2)^4$ and by induction, $$varphi(t)=varphi(t/2^n)^4^n,ngeqslant 1.$$
Using Taylor expansions we see that for large $n$,
$$
varphi(t) = left(1-frac12left(frac t2^nright)^2+oleft(left(frac t2^nright)^2right)right)^4^nstackrelntoinftylongrightarrow expleft(-frac12 t^2right)
$$
and hence $X$ and $Y$ have standard normal distribution.
answered Aug 15 at 10:21
Math1000
18.5k31644
1
For the part to show that $gamma(t)=1+o(t^2)$ you can actually differentiate it once and plug in zero to show that it does not have t term.
– Kenneth Nye
Aug 15 at 10:27
add a comment |Â
up vote
2
down vote
accepted
From independence we have $varphi_X+Y+(X-Y)=varphi_2X$ and thus
$$
varphi(2t) = varphi(t)^2(varphi(t)varphi(-t)) = varphi(t)^3varphi(-t).
$$
Set $gamma(t) = varphi(t)/varphi(-t)$, then
$$
gamma(2t) = fracvarphi(2t)varphi(-2t) = fracvarphi(t)^3varphi(-t)varphi(-t)^3varphi(t) =gamma(t)^2.
$$
It follows that $gamma(t)=gamma(t/2)^2$ and by induction, $gamma(t) = gamma(t/2^n)^2^n$ for nonnegative integers $n$. Moreover, $varphi'(0)=imathbb E[X]=0$ and $varphi''(0)=i^2mathbb E[X^2] = -1$, so as $hto0$ we have by Taylor expansions
$$
gamma(h) = fracvarphi(h)varphi(-h) = frac1-frac12h^2+o(h^2)1-frac12h^2+o(h^2) = 1 + o(h)^2.
$$
It follows that for large $n$, $$gamma(t) = (1+o(t^2/2^2n)^2^nstackrelntoinftylongrightarrow1,$$
and so $gammaequiv1$, yielding $varphi(t)=varphi(-t)$. Therefore $varphi(2t)=varphi(t)^4$, hence $varphi(t)=varphi(t/2)^4$ and by induction, $$varphi(t)=varphi(t/2^n)^4^n,ngeqslant 1.$$
Using Taylor expansions we see that for large $n$,
$$
varphi(t) = left(1-frac12left(frac t2^nright)^2+oleft(left(frac t2^nright)^2right)right)^4^nstackrelntoinftylongrightarrow expleft(-frac12 t^2right)
$$
and hence $X$ and $Y$ have standard normal distribution.
answered Aug 15 at 10:21
Math1000
18.5k31644
1
For the part to show that $gamma(t)=1+o(t^2)$ you can actually differentiate it once and plug in zero to show that it does not have t term.
– Kenneth Nye
Aug 15 at 10:27
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
From independence we have $varphi_X+Y+(X-Y)=varphi_2X$ and thus
$$
varphi(2t) = varphi(t)^2(varphi(t)varphi(-t)) = varphi(t)^3varphi(-t).
$$
Set $gamma(t) = varphi(t)/varphi(-t)$, then
$$
gamma(2t) = fracvarphi(2t)varphi(-2t) = fracvarphi(t)^3varphi(-t)varphi(-t)^3varphi(t) =gamma(t)^2.
$$
It follows that $gamma(t)=gamma(t/2)^2$ and by induction, $gamma(t) = gamma(t/2^n)^2^n$ for nonnegative integers $n$. Moreover, $varphi'(0)=imathbb E[X]=0$ and $varphi''(0)=i^2mathbb E[X^2] = -1$, so as $hto0$ we have by Taylor expansions
$$
gamma(h) = fracvarphi(h)varphi(-h) = frac1-frac12h^2+o(h^2)1-frac12h^2+o(h^2) = 1 + o(h)^2.
$$
It follows that for large $n$, $$gamma(t) = (1+o(t^2/2^2n)^2^nstackrelntoinftylongrightarrow1,$$
and so $gammaequiv1$, yielding $varphi(t)=varphi(-t)$. Therefore $varphi(2t)=varphi(t)^4$, hence $varphi(t)=varphi(t/2)^4$ and by induction, $$varphi(t)=varphi(t/2^n)^4^n,ngeqslant 1.$$
Using Taylor expansions we see that for large $n$,
$$
varphi(t) = left(1-frac12left(frac t2^nright)^2+oleft(left(frac t2^nright)^2right)right)^4^nstackrelntoinftylongrightarrow expleft(-frac12 t^2right)
$$
and hence $X$ and $Y$ have standard normal distribution.
answered Aug 15 at 10:21
Math1000
18.5k31644
From independence we have $varphi_X+Y+(X-Y)=varphi_2X$ and thus
$$
varphi(2t) = varphi(t)^2(varphi(t)varphi(-t)) = varphi(t)^3varphi(-t).
$$
Set $gamma(t) = varphi(t)/varphi(-t)$, then
$$
gamma(2t) = fracvarphi(2t)varphi(-2t) = fracvarphi(t)^3varphi(-t)varphi(-t)^3varphi(t) =gamma(t)^2.
$$
It follows that $gamma(t)=gamma(t/2)^2$ and by induction, $gamma(t) = gamma(t/2^n)^2^n$ for nonnegative integers $n$. Moreover, $varphi'(0)=imathbb E[X]=0$ and $varphi''(0)=i^2mathbb E[X^2] = -1$, so as $hto0$ we have by Taylor expansions
$$
gamma(h) = fracvarphi(h)varphi(-h) = frac1-frac12h^2+o(h^2)1-frac12h^2+o(h^2) = 1 + o(h)^2.
$$
It follows that for large $n$, $$gamma(t) = (1+o(t^2/2^2n)^2^nstackrelntoinftylongrightarrow1,$$
and so $gammaequiv1$, yielding $varphi(t)=varphi(-t)$. Therefore $varphi(2t)=varphi(t)^4$, hence $varphi(t)=varphi(t/2)^4$ and by induction, $$varphi(t)=varphi(t/2^n)^4^n,ngeqslant 1.$$
Using Taylor expansions we see that for large $n$,
$$
varphi(t) = left(1-frac12left(frac t2^nright)^2+oleft(left(frac t2^nright)^2right)right)^4^nstackrelntoinftylongrightarrow expleft(-frac12 t^2right)
$$
and hence $X$ and $Y$ have standard normal distribution.
answered Aug 15 at 10:21
Math1000
18.5k31644
answered Aug 15 at 10:21
Math1000
18.5k31644
answered Aug 15 at 10:21
Math1000
18.5k31644
answered Aug 15 at 10:21
Math1000
18.5k31644
18.5k31644
1
For the part to show that $gamma(t)=1+o(t^2)$ you can actually differentiate it once and plug in zero to show that it does not have t term.
– Kenneth Nye
Aug 15 at 10:27
add a comment |Â
1
For the part to show that $gamma(t)=1+o(t^2)$ you can actually differentiate it once and plug in zero to show that it does not have t term.
– Kenneth Nye
Aug 15 at 10:27
1
1
For the part to show that $gamma(t)=1+o(t^2)$ you can actually differentiate it once and plug in zero to show that it does not have t term.
– Kenneth Nye
Aug 15 at 10:27
For the part to show that $gamma(t)=1+o(t^2)$ you can actually differentiate it once and plug in zero to show that it does not have t term.
– Kenneth Nye
Aug 15 at 10:27
add a comment |Â
up vote
2
down vote
The substitution $gamma(t)$ has the following properties:$$gamma(t) = phi(t)/phi(-t)=fracphi^3(t/2)phi(-t/2)phi(t/2)phi^3(-t/2)=gamma^2(t/2)$$ therefore,$$gamma(t)=lim_ntoinftygamma^2^n(t/2^n)$$
It is easy to show that $$gamma(0)=1$$
$$gamma'(0)=0$$ That means the maclaurin expansion as $tto0$ will be $$gamma(t)=1+o(t^2) $$ thus $$gamma(t)=lim_ntoinftyleft[1+o(t^2/4^n)right]^2^n=1$$ Hence, $$phi(t)=phi(-t)$$ And further that $$phi(2t)=phi^4(t)$$ Therefore, $$phi(t)=phi^4^n(t/2^n)$$ We know that $$phi(t)=1+E(X)ti-fracE(X^2)2t^2+o(t^2)=1-fract^22+o(t^2)$$ then we get that long expression. By taking limit on it we get $$phi(t)=lim_ntoinftyleft1-frac12fract^24^nright^4^n=e^-frac12t^2$$
answered Aug 15 at 10:21
Kenneth Nye
16210
Very nice work! :-)
– Ant
Aug 15 at 10:23
add a comment |Â
up vote
2
down vote
The substitution $gamma(t)$ has the following properties:$$gamma(t) = phi(t)/phi(-t)=fracphi^3(t/2)phi(-t/2)phi(t/2)phi^3(-t/2)=gamma^2(t/2)$$ therefore,$$gamma(t)=lim_ntoinftygamma^2^n(t/2^n)$$
It is easy to show that $$gamma(0)=1$$
$$gamma'(0)=0$$ That means the maclaurin expansion as $tto0$ will be $$gamma(t)=1+o(t^2) $$ thus $$gamma(t)=lim_ntoinftyleft[1+o(t^2/4^n)right]^2^n=1$$ Hence, $$phi(t)=phi(-t)$$ And further that $$phi(2t)=phi^4(t)$$ Therefore, $$phi(t)=phi^4^n(t/2^n)$$ We know that $$phi(t)=1+E(X)ti-fracE(X^2)2t^2+o(t^2)=1-fract^22+o(t^2)$$ then we get that long expression. By taking limit on it we get $$phi(t)=lim_ntoinftyleft1-frac12fract^24^nright^4^n=e^-frac12t^2$$
answered Aug 15 at 10:21
Kenneth Nye
16210
Very nice work! :-)
– Ant
Aug 15 at 10:23
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The substitution $gamma(t)$ has the following properties:$$gamma(t) = phi(t)/phi(-t)=fracphi^3(t/2)phi(-t/2)phi(t/2)phi^3(-t/2)=gamma^2(t/2)$$ therefore,$$gamma(t)=lim_ntoinftygamma^2^n(t/2^n)$$
It is easy to show that $$gamma(0)=1$$
$$gamma'(0)=0$$ That means the maclaurin expansion as $tto0$ will be $$gamma(t)=1+o(t^2) $$ thus $$gamma(t)=lim_ntoinftyleft[1+o(t^2/4^n)right]^2^n=1$$ Hence, $$phi(t)=phi(-t)$$ And further that $$phi(2t)=phi^4(t)$$ Therefore, $$phi(t)=phi^4^n(t/2^n)$$ We know that $$phi(t)=1+E(X)ti-fracE(X^2)2t^2+o(t^2)=1-fract^22+o(t^2)$$ then we get that long expression. By taking limit on it we get $$phi(t)=lim_ntoinftyleft1-frac12fract^24^nright^4^n=e^-frac12t^2$$
answered Aug 15 at 10:21
Kenneth Nye
16210
The substitution $gamma(t)$ has the following properties:$$gamma(t) = phi(t)/phi(-t)=fracphi^3(t/2)phi(-t/2)phi(t/2)phi^3(-t/2)=gamma^2(t/2)$$ therefore,$$gamma(t)=lim_ntoinftygamma^2^n(t/2^n)$$
It is easy to show that $$gamma(0)=1$$
$$gamma'(0)=0$$ That means the maclaurin expansion as $tto0$ will be $$gamma(t)=1+o(t^2) $$ thus $$gamma(t)=lim_ntoinftyleft[1+o(t^2/4^n)right]^2^n=1$$ Hence, $$phi(t)=phi(-t)$$ And further that $$phi(2t)=phi^4(t)$$ Therefore, $$phi(t)=phi^4^n(t/2^n)$$ We know that $$phi(t)=1+E(X)ti-fracE(X^2)2t^2+o(t^2)=1-fract^22+o(t^2)$$ then we get that long expression. By taking limit on it we get $$phi(t)=lim_ntoinftyleft1-frac12fract^24^nright^4^n=e^-frac12t^2$$
answered Aug 15 at 10:21
Kenneth Nye
16210
answered Aug 15 at 10:21
Kenneth Nye
16210
answered Aug 15 at 10:21
Kenneth Nye
16210
answered Aug 15 at 10:21
Kenneth Nye
16210
16210
Very nice work! :-)
– Ant
Aug 15 at 10:23
add a comment |Â
Very nice work! :-)
– Ant
Aug 15 at 10:23
Very nice work! :-)
– Ant
Aug 15 at 10:23
Very nice work! :-)
– Ant
Aug 15 at 10:23
add a comment |Â
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1
Not sure yet how to solve it but you can start from the observation that for every positive integer $n$, $gamma(2^n t) = gamma(t)^2n$
– Ant
Aug 15 at 9:08
@Ant Just realize how you get that observation and remember it appeared somewhere in the textbook. I will see how...
– Kenneth Nye
Aug 15 at 9:19
For the case of a symmetric distribution this is fairly easy. Expanding the characteristic function around $0$ and taking the limit shows that the common distribution is normal. How one can use $gamma $ to handle non-symmetric case beats me.
– Kavi Rama Murthy
Aug 15 at 9:31
@Ant I figured out the answer. Thanks.
– Kenneth Nye
Aug 15 at 10:21