Is this a Cauchy random variable?
$begingroup$
Suppose $X$ and $Y$ are independent $n(0,1)$ random variables.
How do I find $P(X^2+Y^2leq 1)$ and $P(X^2leq 1)$, after verifying that $X^2$ is distributed $chi_{1}^{2}$?
My attempt:
$$P(X^2+Y^2leq 1)= int_{-infty}^{infty}int_{-infty}^sqrt{1-y^2} frac{1}{2pi} e^frac{-x^2+y^2}{2} dxdy $$
How do I proceed from it?
probability probability-theory statistics
asked Dec 6 '18 at 2:25
LadyLady
1198
$endgroup$
add a comment |
$begingroup$
Suppose $X$ and $Y$ are independent $n(0,1)$ random variables.
How do I find $P(X^2+Y^2leq 1)$ and $P(X^2leq 1)$, after verifying that $X^2$ is distributed $chi_{1}^{2}$?
My attempt:
$$P(X^2+Y^2leq 1)= int_{-infty}^{infty}int_{-infty}^sqrt{1-y^2} frac{1}{2pi} e^frac{-x^2+y^2}{2} dxdy $$
How do I proceed from it?
probability probability-theory statistics
asked Dec 6 '18 at 2:25
LadyLady
1198
$endgroup$
$begingroup$
Should your title be "Is this a chi-squared random variable?"?
$endgroup$
– Henry
Dec 6 '18 at 10:29
$begingroup$
$X^2 sim CHISQ(1), P(X^2 < 1) = P(-1 < X < 1) = 0.6826895$ from R wherepchisandpnormare CDFs or the 2 relevant dist'ns: codepchisq(1, 1)anddiff(pnorm(c(-1,1)))both return 0.6826895. Also $Q=X^2 + Y^2 sim CHISQ(2),$ perhaps most easily proved using MGFs. $P(Q le 1) $ can be found in R frompchisq(1, 2), which returns 0.3934693.
$endgroup$
– BruceET
Dec 8 '18 at 2:28
$begingroup$
You might get an aprx value for $P(Q le 1)$ from looking at a printed table of the chi-sq dist'n with DF=2. Not all printed tables give enough detail to be helpful, but some do. // But if you haven't started using R, you can get it for free and with very little trouble online. Start by learning how the probability functions work:pnorm,dbinom,runif, etc. Just use the bits of R that are immediately helpful to you. There is good user-group help online. You'll never learn all of it. If you did, you'd be the first.
$endgroup$
– BruceET
Dec 8 '18 at 2:37
add a comment |
$begingroup$
Suppose $X$ and $Y$ are independent $n(0,1)$ random variables.
How do I find $P(X^2+Y^2leq 1)$ and $P(X^2leq 1)$, after verifying that $X^2$ is distributed $chi_{1}^{2}$?
My attempt:
$$P(X^2+Y^2leq 1)= int_{-infty}^{infty}int_{-infty}^sqrt{1-y^2} frac{1}{2pi} e^frac{-x^2+y^2}{2} dxdy $$
How do I proceed from it?
probability probability-theory statistics
asked Dec 6 '18 at 2:25
LadyLady
1198
$endgroup$
Suppose $X$ and $Y$ are independent $n(0,1)$ random variables.
How do I find $P(X^2+Y^2leq 1)$ and $P(X^2leq 1)$, after verifying that $X^2$ is distributed $chi_{1}^{2}$?
My attempt:
$$P(X^2+Y^2leq 1)= int_{-infty}^{infty}int_{-infty}^sqrt{1-y^2} frac{1}{2pi} e^frac{-x^2+y^2}{2} dxdy $$
How do I proceed from it?
probability probability-theory statistics
probability probability-theory statistics
asked Dec 6 '18 at 2:25
LadyLady
1198
asked Dec 6 '18 at 2:25
LadyLady
1198
asked Dec 6 '18 at 2:25
LadyLady
1198
asked Dec 6 '18 at 2:25
LadyLady
1198
asked Dec 6 '18 at 2:25
LadyLady
1198
1198
$begingroup$
Should your title be "Is this a chi-squared random variable?"?
$endgroup$
– Henry
Dec 6 '18 at 10:29
$begingroup$
$X^2 sim CHISQ(1), P(X^2 < 1) = P(-1 < X < 1) = 0.6826895$ from R wherepchisandpnormare CDFs or the 2 relevant dist'ns: codepchisq(1, 1)anddiff(pnorm(c(-1,1)))both return 0.6826895. Also $Q=X^2 + Y^2 sim CHISQ(2),$ perhaps most easily proved using MGFs. $P(Q le 1) $ can be found in R frompchisq(1, 2), which returns 0.3934693.
$endgroup$
– BruceET
Dec 8 '18 at 2:28
$begingroup$
You might get an aprx value for $P(Q le 1)$ from looking at a printed table of the chi-sq dist'n with DF=2. Not all printed tables give enough detail to be helpful, but some do. // But if you haven't started using R, you can get it for free and with very little trouble online. Start by learning how the probability functions work:pnorm,dbinom,runif, etc. Just use the bits of R that are immediately helpful to you. There is good user-group help online. You'll never learn all of it. If you did, you'd be the first.
$endgroup$
– BruceET
Dec 8 '18 at 2:37
add a comment |
$begingroup$
Should your title be "Is this a chi-squared random variable?"?
$endgroup$
– Henry
Dec 6 '18 at 10:29
$begingroup$
$X^2 sim CHISQ(1), P(X^2 < 1) = P(-1 < X < 1) = 0.6826895$ from R wherepchisandpnormare CDFs or the 2 relevant dist'ns: codepchisq(1, 1)anddiff(pnorm(c(-1,1)))both return 0.6826895. Also $Q=X^2 + Y^2 sim CHISQ(2),$ perhaps most easily proved using MGFs. $P(Q le 1) $ can be found in R frompchisq(1, 2), which returns 0.3934693.
$endgroup$
– BruceET
Dec 8 '18 at 2:28
$begingroup$
You might get an aprx value for $P(Q le 1)$ from looking at a printed table of the chi-sq dist'n with DF=2. Not all printed tables give enough detail to be helpful, but some do. // But if you haven't started using R, you can get it for free and with very little trouble online. Start by learning how the probability functions work:pnorm,dbinom,runif, etc. Just use the bits of R that are immediately helpful to you. There is good user-group help online. You'll never learn all of it. If you did, you'd be the first.
$endgroup$
– BruceET
Dec 8 '18 at 2:37
$begingroup$
Should your title be "Is this a chi-squared random variable?"?
$endgroup$
– Henry
Dec 6 '18 at 10:29
$begingroup$
Should your title be "Is this a chi-squared random variable?"?
$endgroup$
– Henry
Dec 6 '18 at 10:29
$begingroup$
$X^2 sim CHISQ(1), P(X^2 < 1) = P(-1 < X < 1) = 0.6826895$ from R where
pchis and pnorm are CDFs or the 2 relevant dist'ns: code pchisq(1, 1) and diff(pnorm(c(-1,1))) both return 0.6826895. Also $Q=X^2 + Y^2 sim CHISQ(2),$ perhaps most easily proved using MGFs. $P(Q le 1) $ can be found in R from pchisq(1, 2), which returns 0.3934693.$endgroup$
– BruceET
Dec 8 '18 at 2:28
$begingroup$
$X^2 sim CHISQ(1), P(X^2 < 1) = P(-1 < X < 1) = 0.6826895$ from R where
pchis and pnorm are CDFs or the 2 relevant dist'ns: code pchisq(1, 1) and diff(pnorm(c(-1,1))) both return 0.6826895. Also $Q=X^2 + Y^2 sim CHISQ(2),$ perhaps most easily proved using MGFs. $P(Q le 1) $ can be found in R from pchisq(1, 2), which returns 0.3934693.$endgroup$
– BruceET
Dec 8 '18 at 2:28
$begingroup$
You might get an aprx value for $P(Q le 1)$ from looking at a printed table of the chi-sq dist'n with DF=2. Not all printed tables give enough detail to be helpful, but some do. // But if you haven't started using R, you can get it for free and with very little trouble online. Start by learning how the probability functions work:
pnorm, dbinom, runif, etc. Just use the bits of R that are immediately helpful to you. There is good user-group help online. You'll never learn all of it. If you did, you'd be the first.$endgroup$
– BruceET
Dec 8 '18 at 2:37
$begingroup$
You might get an aprx value for $P(Q le 1)$ from looking at a printed table of the chi-sq dist'n with DF=2. Not all printed tables give enough detail to be helpful, but some do. // But if you haven't started using R, you can get it for free and with very little trouble online. Start by learning how the probability functions work:
pnorm, dbinom, runif, etc. Just use the bits of R that are immediately helpful to you. There is good user-group help online. You'll never learn all of it. If you did, you'd be the first.$endgroup$
– BruceET
Dec 8 '18 at 2:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The crux of your question appears to be: "How do I show that if $X sim N(0, 1)$, then $X^2 sim chi^2(1)$?"
Here's the gist: Let $f(x), F(x)$ be the density and cdf of a standard normal variable, respectively. (Note that $f(x) = frac{1}{sqrt{2 pi}} e^{-x^2/2}$, but that $F(x)$ doesn't have a clean closed-form expression.) We want to know more about the variable $U = X^2$, so call its density and cdf by $g(x)$ and $G(x)$, respectively.
First, note that for $x > 0$,
begin{align*}G(x) &= mathbb P(U leq x) \ &= mathbb P(X^2 leq x) \ &= mathbb P(-sqrt x leq X leq sqrt x) \ &= F(sqrt x) - F(-sqrt x).
end{align*}
It follows that
begin{align*}
g(x) &= G'(x) \ &= frac{textrm d}{textrm d x} [F(sqrt x) - F(-sqrt x)] \
&= f(sqrt x) cdot frac{1}{2 sqrt x} + f(-sqrt x) cdot frac{1}{2 sqrt x}
end{align*}
for all $x > 0$, and if you evaluate that, it happens to be exactly the density of a $chi^2$ variable with 1 df.
There are lots of details in the steps above, but I wasn't sure about your background in the material, so let me know if there are steps that need clarification.
answered Dec 6 '18 at 4:19
Aaron MontgomeryAaron Montgomery
4,732523
$endgroup$
add a comment |
$begingroup$
Since $X,Ysim N(0,1)$, then $X^2$ and $Y^2$ are chi-squared with 1 degree of freedom each.
Since $X^2,Y^2simchi^2(1)$ and they are independent, then $X^2+Y^2simchi^2(2)$.
So to compute those probabilities, use a chi-squared density (or R).
answered Dec 6 '18 at 2:34
gd1035gd1035
4571210
$endgroup$
$begingroup$
How do I find $P(X^2leq 1)$, after verifying that $X^2$ is distributed $chi_{1}^{2}$?
$endgroup$
– Lady
Dec 6 '18 at 2:40
$begingroup$
Using a chi-squared density with 1 degree of freedom. The density is $f_{X^2}(x)=frac{1}{Gamma(1/2)2^{1/2}}x^{-1/2}e^{-x/2}$. Or if you are using r,pchisq(1,1)
$endgroup$
– gd1035
Dec 6 '18 at 2:44
$begingroup$
But how do I show that $X^2$ is distributed $chi_{1}^{2}$?
$endgroup$
– Lady
Dec 6 '18 at 2:48
add a comment |
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2 Answers
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2 Answers
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active
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oldest
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$begingroup$
The crux of your question appears to be: "How do I show that if $X sim N(0, 1)$, then $X^2 sim chi^2(1)$?"
Here's the gist: Let $f(x), F(x)$ be the density and cdf of a standard normal variable, respectively. (Note that $f(x) = frac{1}{sqrt{2 pi}} e^{-x^2/2}$, but that $F(x)$ doesn't have a clean closed-form expression.) We want to know more about the variable $U = X^2$, so call its density and cdf by $g(x)$ and $G(x)$, respectively.
First, note that for $x > 0$,
begin{align*}G(x) &= mathbb P(U leq x) \ &= mathbb P(X^2 leq x) \ &= mathbb P(-sqrt x leq X leq sqrt x) \ &= F(sqrt x) - F(-sqrt x).
end{align*}
It follows that
begin{align*}
g(x) &= G'(x) \ &= frac{textrm d}{textrm d x} [F(sqrt x) - F(-sqrt x)] \
&= f(sqrt x) cdot frac{1}{2 sqrt x} + f(-sqrt x) cdot frac{1}{2 sqrt x}
end{align*}
for all $x > 0$, and if you evaluate that, it happens to be exactly the density of a $chi^2$ variable with 1 df.
There are lots of details in the steps above, but I wasn't sure about your background in the material, so let me know if there are steps that need clarification.
answered Dec 6 '18 at 4:19
Aaron MontgomeryAaron Montgomery
4,732523
$endgroup$
add a comment |
$begingroup$
The crux of your question appears to be: "How do I show that if $X sim N(0, 1)$, then $X^2 sim chi^2(1)$?"
Here's the gist: Let $f(x), F(x)$ be the density and cdf of a standard normal variable, respectively. (Note that $f(x) = frac{1}{sqrt{2 pi}} e^{-x^2/2}$, but that $F(x)$ doesn't have a clean closed-form expression.) We want to know more about the variable $U = X^2$, so call its density and cdf by $g(x)$ and $G(x)$, respectively.
First, note that for $x > 0$,
begin{align*}G(x) &= mathbb P(U leq x) \ &= mathbb P(X^2 leq x) \ &= mathbb P(-sqrt x leq X leq sqrt x) \ &= F(sqrt x) - F(-sqrt x).
end{align*}
It follows that
begin{align*}
g(x) &= G'(x) \ &= frac{textrm d}{textrm d x} [F(sqrt x) - F(-sqrt x)] \
&= f(sqrt x) cdot frac{1}{2 sqrt x} + f(-sqrt x) cdot frac{1}{2 sqrt x}
end{align*}
for all $x > 0$, and if you evaluate that, it happens to be exactly the density of a $chi^2$ variable with 1 df.
There are lots of details in the steps above, but I wasn't sure about your background in the material, so let me know if there are steps that need clarification.
answered Dec 6 '18 at 4:19
Aaron MontgomeryAaron Montgomery
4,732523
$endgroup$
add a comment |
$begingroup$
The crux of your question appears to be: "How do I show that if $X sim N(0, 1)$, then $X^2 sim chi^2(1)$?"
Here's the gist: Let $f(x), F(x)$ be the density and cdf of a standard normal variable, respectively. (Note that $f(x) = frac{1}{sqrt{2 pi}} e^{-x^2/2}$, but that $F(x)$ doesn't have a clean closed-form expression.) We want to know more about the variable $U = X^2$, so call its density and cdf by $g(x)$ and $G(x)$, respectively.
First, note that for $x > 0$,
begin{align*}G(x) &= mathbb P(U leq x) \ &= mathbb P(X^2 leq x) \ &= mathbb P(-sqrt x leq X leq sqrt x) \ &= F(sqrt x) - F(-sqrt x).
end{align*}
It follows that
begin{align*}
g(x) &= G'(x) \ &= frac{textrm d}{textrm d x} [F(sqrt x) - F(-sqrt x)] \
&= f(sqrt x) cdot frac{1}{2 sqrt x} + f(-sqrt x) cdot frac{1}{2 sqrt x}
end{align*}
for all $x > 0$, and if you evaluate that, it happens to be exactly the density of a $chi^2$ variable with 1 df.
There are lots of details in the steps above, but I wasn't sure about your background in the material, so let me know if there are steps that need clarification.
answered Dec 6 '18 at 4:19
Aaron MontgomeryAaron Montgomery
4,732523
$endgroup$
The crux of your question appears to be: "How do I show that if $X sim N(0, 1)$, then $X^2 sim chi^2(1)$?"
Here's the gist: Let $f(x), F(x)$ be the density and cdf of a standard normal variable, respectively. (Note that $f(x) = frac{1}{sqrt{2 pi}} e^{-x^2/2}$, but that $F(x)$ doesn't have a clean closed-form expression.) We want to know more about the variable $U = X^2$, so call its density and cdf by $g(x)$ and $G(x)$, respectively.
First, note that for $x > 0$,
begin{align*}G(x) &= mathbb P(U leq x) \ &= mathbb P(X^2 leq x) \ &= mathbb P(-sqrt x leq X leq sqrt x) \ &= F(sqrt x) - F(-sqrt x).
end{align*}
It follows that
begin{align*}
g(x) &= G'(x) \ &= frac{textrm d}{textrm d x} [F(sqrt x) - F(-sqrt x)] \
&= f(sqrt x) cdot frac{1}{2 sqrt x} + f(-sqrt x) cdot frac{1}{2 sqrt x}
end{align*}
for all $x > 0$, and if you evaluate that, it happens to be exactly the density of a $chi^2$ variable with 1 df.
There are lots of details in the steps above, but I wasn't sure about your background in the material, so let me know if there are steps that need clarification.
answered Dec 6 '18 at 4:19
Aaron MontgomeryAaron Montgomery
4,732523
answered Dec 6 '18 at 4:19
Aaron MontgomeryAaron Montgomery
4,732523
answered Dec 6 '18 at 4:19
Aaron MontgomeryAaron Montgomery
4,732523
answered Dec 6 '18 at 4:19
Aaron MontgomeryAaron Montgomery
4,732523
4,732523
add a comment |
add a comment |
$begingroup$
Since $X,Ysim N(0,1)$, then $X^2$ and $Y^2$ are chi-squared with 1 degree of freedom each.
Since $X^2,Y^2simchi^2(1)$ and they are independent, then $X^2+Y^2simchi^2(2)$.
So to compute those probabilities, use a chi-squared density (or R).
answered Dec 6 '18 at 2:34
gd1035gd1035
4571210
$endgroup$
$begingroup$
How do I find $P(X^2leq 1)$, after verifying that $X^2$ is distributed $chi_{1}^{2}$?
$endgroup$
– Lady
Dec 6 '18 at 2:40
$begingroup$
Using a chi-squared density with 1 degree of freedom. The density is $f_{X^2}(x)=frac{1}{Gamma(1/2)2^{1/2}}x^{-1/2}e^{-x/2}$. Or if you are using r,pchisq(1,1)
$endgroup$
– gd1035
Dec 6 '18 at 2:44
$begingroup$
But how do I show that $X^2$ is distributed $chi_{1}^{2}$?
$endgroup$
– Lady
Dec 6 '18 at 2:48
add a comment |
$begingroup$
Since $X,Ysim N(0,1)$, then $X^2$ and $Y^2$ are chi-squared with 1 degree of freedom each.
Since $X^2,Y^2simchi^2(1)$ and they are independent, then $X^2+Y^2simchi^2(2)$.
So to compute those probabilities, use a chi-squared density (or R).
answered Dec 6 '18 at 2:34
gd1035gd1035
4571210
$endgroup$
$begingroup$
How do I find $P(X^2leq 1)$, after verifying that $X^2$ is distributed $chi_{1}^{2}$?
$endgroup$
– Lady
Dec 6 '18 at 2:40
$begingroup$
Using a chi-squared density with 1 degree of freedom. The density is $f_{X^2}(x)=frac{1}{Gamma(1/2)2^{1/2}}x^{-1/2}e^{-x/2}$. Or if you are using r,pchisq(1,1)
$endgroup$
– gd1035
Dec 6 '18 at 2:44
$begingroup$
But how do I show that $X^2$ is distributed $chi_{1}^{2}$?
$endgroup$
– Lady
Dec 6 '18 at 2:48
add a comment |
$begingroup$
Since $X,Ysim N(0,1)$, then $X^2$ and $Y^2$ are chi-squared with 1 degree of freedom each.
Since $X^2,Y^2simchi^2(1)$ and they are independent, then $X^2+Y^2simchi^2(2)$.
So to compute those probabilities, use a chi-squared density (or R).
answered Dec 6 '18 at 2:34
gd1035gd1035
4571210
$endgroup$
Since $X,Ysim N(0,1)$, then $X^2$ and $Y^2$ are chi-squared with 1 degree of freedom each.
Since $X^2,Y^2simchi^2(1)$ and they are independent, then $X^2+Y^2simchi^2(2)$.
So to compute those probabilities, use a chi-squared density (or R).
answered Dec 6 '18 at 2:34
gd1035gd1035
4571210
edited Dec 6 '18 at 2:40
answered Dec 6 '18 at 2:34
gd1035gd1035
4571210
answered Dec 6 '18 at 2:34
gd1035gd1035
4571210
answered Dec 6 '18 at 2:34
gd1035gd1035
4571210
4571210
$begingroup$
How do I find $P(X^2leq 1)$, after verifying that $X^2$ is distributed $chi_{1}^{2}$?
$endgroup$
– Lady
Dec 6 '18 at 2:40
$begingroup$
Using a chi-squared density with 1 degree of freedom. The density is $f_{X^2}(x)=frac{1}{Gamma(1/2)2^{1/2}}x^{-1/2}e^{-x/2}$. Or if you are using r,pchisq(1,1)
$endgroup$
– gd1035
Dec 6 '18 at 2:44
$begingroup$
But how do I show that $X^2$ is distributed $chi_{1}^{2}$?
$endgroup$
– Lady
Dec 6 '18 at 2:48
add a comment |
$begingroup$
How do I find $P(X^2leq 1)$, after verifying that $X^2$ is distributed $chi_{1}^{2}$?
$endgroup$
– Lady
Dec 6 '18 at 2:40
$begingroup$
Using a chi-squared density with 1 degree of freedom. The density is $f_{X^2}(x)=frac{1}{Gamma(1/2)2^{1/2}}x^{-1/2}e^{-x/2}$. Or if you are using r,pchisq(1,1)
$endgroup$
– gd1035
Dec 6 '18 at 2:44
$begingroup$
But how do I show that $X^2$ is distributed $chi_{1}^{2}$?
$endgroup$
– Lady
Dec 6 '18 at 2:48
$begingroup$
How do I find $P(X^2leq 1)$, after verifying that $X^2$ is distributed $chi_{1}^{2}$?
$endgroup$
– Lady
Dec 6 '18 at 2:40
$begingroup$
How do I find $P(X^2leq 1)$, after verifying that $X^2$ is distributed $chi_{1}^{2}$?
$endgroup$
– Lady
Dec 6 '18 at 2:40
$begingroup$
Using a chi-squared density with 1 degree of freedom. The density is $f_{X^2}(x)=frac{1}{Gamma(1/2)2^{1/2}}x^{-1/2}e^{-x/2}$. Or if you are using r,
pchisq(1,1)$endgroup$
– gd1035
Dec 6 '18 at 2:44
$begingroup$
Using a chi-squared density with 1 degree of freedom. The density is $f_{X^2}(x)=frac{1}{Gamma(1/2)2^{1/2}}x^{-1/2}e^{-x/2}$. Or if you are using r,
pchisq(1,1)$endgroup$
– gd1035
Dec 6 '18 at 2:44
$begingroup$
But how do I show that $X^2$ is distributed $chi_{1}^{2}$?
$endgroup$
– Lady
Dec 6 '18 at 2:48
$begingroup$
But how do I show that $X^2$ is distributed $chi_{1}^{2}$?
$endgroup$
– Lady
Dec 6 '18 at 2:48
add a comment |
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$begingroup$
Should your title be "Is this a chi-squared random variable?"?
$endgroup$
– Henry
Dec 6 '18 at 10:29
$begingroup$
$X^2 sim CHISQ(1), P(X^2 < 1) = P(-1 < X < 1) = 0.6826895$ from R where
pchisandpnormare CDFs or the 2 relevant dist'ns: codepchisq(1, 1)anddiff(pnorm(c(-1,1)))both return 0.6826895. Also $Q=X^2 + Y^2 sim CHISQ(2),$ perhaps most easily proved using MGFs. $P(Q le 1) $ can be found in R frompchisq(1, 2), which returns 0.3934693.$endgroup$
– BruceET
Dec 8 '18 at 2:28
$begingroup$
You might get an aprx value for $P(Q le 1)$ from looking at a printed table of the chi-sq dist'n with DF=2. Not all printed tables give enough detail to be helpful, but some do. // But if you haven't started using R, you can get it for free and with very little trouble online. Start by learning how the probability functions work:
pnorm,dbinom,runif, etc. Just use the bits of R that are immediately helpful to you. There is good user-group help online. You'll never learn all of it. If you did, you'd be the first.$endgroup$
– BruceET
Dec 8 '18 at 2:37