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Is this a typo or a delicate point to understand?
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Suppose $X$ is a discrete random variable with distribution $F(t) = P(X leq t)$. We know $P(X < t) = lim_s to t^- F(t) $. In trying to find a probability like $P(1/2 < X < 3/2) $, I would do
$$ P(X<3/2) - P(X<1/2) = lim_x to 3/2^- F(t) - lim_x to 1/2^- F(t) $$
However, professor wrote
$$ P(1/2 < X < 3/2) = P(X<3/2) - P(X leq 1/2) = P(X<3/2) - F(1/2) $$
is this a typo? or I am misunderstanding the concept?
probability
asked 4 hours ago
Jimmy Sabater
1,50014
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up vote
2
down vote
favorite
Suppose $X$ is a discrete random variable with distribution $F(t) = P(X leq t)$. We know $P(X < t) = lim_s to t^- F(t) $. In trying to find a probability like $P(1/2 < X < 3/2) $, I would do
$$ P(X<3/2) - P(X<1/2) = lim_x to 3/2^- F(t) - lim_x to 1/2^- F(t) $$
However, professor wrote
$$ P(1/2 < X < 3/2) = P(X<3/2) - P(X leq 1/2) = P(X<3/2) - F(1/2) $$
is this a typo? or I am misunderstanding the concept?
probability
asked 4 hours ago
Jimmy Sabater
1,50014
We have $$X<3/2 = Xleq 1/2 cup 1/2<X<3/2$$ where the right-hand-side is a union of disjoint events, and so $$P[X<3/2] = P[Xleq 1/2] + P[1/2<X<3/2]$$
– Michael
4 hours ago
Note that $P[a < X le b] = F(b)-F(a)$.
– copper.hat
4 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose $X$ is a discrete random variable with distribution $F(t) = P(X leq t)$. We know $P(X < t) = lim_s to t^- F(t) $. In trying to find a probability like $P(1/2 < X < 3/2) $, I would do
$$ P(X<3/2) - P(X<1/2) = lim_x to 3/2^- F(t) - lim_x to 1/2^- F(t) $$
However, professor wrote
$$ P(1/2 < X < 3/2) = P(X<3/2) - P(X leq 1/2) = P(X<3/2) - F(1/2) $$
is this a typo? or I am misunderstanding the concept?
probability
asked 4 hours ago
Jimmy Sabater
1,50014
Suppose $X$ is a discrete random variable with distribution $F(t) = P(X leq t)$. We know $P(X < t) = lim_s to t^- F(t) $. In trying to find a probability like $P(1/2 < X < 3/2) $, I would do
$$ P(X<3/2) - P(X<1/2) = lim_x to 3/2^- F(t) - lim_x to 1/2^- F(t) $$
However, professor wrote
$$ P(1/2 < X < 3/2) = P(X<3/2) - P(X leq 1/2) = P(X<3/2) - F(1/2) $$
is this a typo? or I am misunderstanding the concept?
probability
probability
asked 4 hours ago
Jimmy Sabater
1,50014
asked 4 hours ago
Jimmy Sabater
1,50014
asked 4 hours ago
Jimmy Sabater
1,50014
asked 4 hours ago
Jimmy Sabater
1,50014
asked 4 hours ago
Jimmy Sabater
1,50014
1,50014
We have $$X<3/2 = Xleq 1/2 cup 1/2<X<3/2$$ where the right-hand-side is a union of disjoint events, and so $$P[X<3/2] = P[Xleq 1/2] + P[1/2<X<3/2]$$
– Michael
4 hours ago
Note that $P[a < X le b] = F(b)-F(a)$.
– copper.hat
4 hours ago
add a comment |Â
We have $$X<3/2 = Xleq 1/2 cup 1/2<X<3/2$$ where the right-hand-side is a union of disjoint events, and so $$P[X<3/2] = P[Xleq 1/2] + P[1/2<X<3/2]$$
– Michael
4 hours ago
Note that $P[a < X le b] = F(b)-F(a)$.
– copper.hat
4 hours ago
We have $$X<3/2 = Xleq 1/2 cup 1/2<X<3/2$$ where the right-hand-side is a union of disjoint events, and so $$P[X<3/2] = P[Xleq 1/2] + P[1/2<X<3/2]$$
– Michael
4 hours ago
We have $$X<3/2 = Xleq 1/2 cup 1/2<X<3/2$$ where the right-hand-side is a union of disjoint events, and so $$P[X<3/2] = P[Xleq 1/2] + P[1/2<X<3/2]$$
– Michael
4 hours ago
Note that $P[a < X le b] = F(b)-F(a)$.
– copper.hat
4 hours ago
Note that $P[a < X le b] = F(b)-F(a)$.
– copper.hat
4 hours ago
add a comment |Â
2 Answers
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Not a typo.
The notation $(1/2 < X < 3/2)$ is just a convenient notation for the event (=set) $ X in (1/2,3/2); $. In particular, since the interval $(1/2,3/2)$ can be expressed as $Asetminus B$, where $A=(-infty, 3/2)$ and $B=(-infty, 1/2]$ (note the square bracket at the end), we have
$$P(1/2 < X < 3/2) = P(Asetminus B) = P(X<3/2) - P(Xleqslant 1/2)$$
By definition, the last term is $P(Xleqslant 1/2)=F(1/2)$.
In other words, being strictly between $1/2$ and $3/2$ (that is, $1/2<X<3/2$) means being smaller than $3/2$ (that is, $X<3/2$) but not smaller than or equal to $1/2$ (that is, $X leqslant 1/2$).
If $X=1/2$, then $X$ does not satisfy $1/2 < X < 3/2$.
answered 4 hours ago
Taladris
4,46931732
add a comment |Â
up vote
3
down vote
Your professor's answer is correct
Remember that you are dealing with a discrete random variable, so you can't just assume that the probability of being exactly $1/2$ is zero, so you need to remember to subtract it off.
Your solution finds $P(1/2 le X lt 3/2)$ which isnt necessarily the same for discrete random variables
answered 4 hours ago
WaveX
2,1552720
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Not a typo.
The notation $(1/2 < X < 3/2)$ is just a convenient notation for the event (=set) $ X in (1/2,3/2); $. In particular, since the interval $(1/2,3/2)$ can be expressed as $Asetminus B$, where $A=(-infty, 3/2)$ and $B=(-infty, 1/2]$ (note the square bracket at the end), we have
$$P(1/2 < X < 3/2) = P(Asetminus B) = P(X<3/2) - P(Xleqslant 1/2)$$
By definition, the last term is $P(Xleqslant 1/2)=F(1/2)$.
In other words, being strictly between $1/2$ and $3/2$ (that is, $1/2<X<3/2$) means being smaller than $3/2$ (that is, $X<3/2$) but not smaller than or equal to $1/2$ (that is, $X leqslant 1/2$).
If $X=1/2$, then $X$ does not satisfy $1/2 < X < 3/2$.
answered 4 hours ago
Taladris
4,46931732
add a comment |Â
up vote
3
down vote
accepted
Not a typo.
The notation $(1/2 < X < 3/2)$ is just a convenient notation for the event (=set) $ X in (1/2,3/2); $. In particular, since the interval $(1/2,3/2)$ can be expressed as $Asetminus B$, where $A=(-infty, 3/2)$ and $B=(-infty, 1/2]$ (note the square bracket at the end), we have
$$P(1/2 < X < 3/2) = P(Asetminus B) = P(X<3/2) - P(Xleqslant 1/2)$$
By definition, the last term is $P(Xleqslant 1/2)=F(1/2)$.
In other words, being strictly between $1/2$ and $3/2$ (that is, $1/2<X<3/2$) means being smaller than $3/2$ (that is, $X<3/2$) but not smaller than or equal to $1/2$ (that is, $X leqslant 1/2$).
If $X=1/2$, then $X$ does not satisfy $1/2 < X < 3/2$.
answered 4 hours ago
Taladris
4,46931732
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Not a typo.
The notation $(1/2 < X < 3/2)$ is just a convenient notation for the event (=set) $ X in (1/2,3/2); $. In particular, since the interval $(1/2,3/2)$ can be expressed as $Asetminus B$, where $A=(-infty, 3/2)$ and $B=(-infty, 1/2]$ (note the square bracket at the end), we have
$$P(1/2 < X < 3/2) = P(Asetminus B) = P(X<3/2) - P(Xleqslant 1/2)$$
By definition, the last term is $P(Xleqslant 1/2)=F(1/2)$.
In other words, being strictly between $1/2$ and $3/2$ (that is, $1/2<X<3/2$) means being smaller than $3/2$ (that is, $X<3/2$) but not smaller than or equal to $1/2$ (that is, $X leqslant 1/2$).
If $X=1/2$, then $X$ does not satisfy $1/2 < X < 3/2$.
answered 4 hours ago
Taladris
4,46931732
Not a typo.
The notation $(1/2 < X < 3/2)$ is just a convenient notation for the event (=set) $ X in (1/2,3/2); $. In particular, since the interval $(1/2,3/2)$ can be expressed as $Asetminus B$, where $A=(-infty, 3/2)$ and $B=(-infty, 1/2]$ (note the square bracket at the end), we have
$$P(1/2 < X < 3/2) = P(Asetminus B) = P(X<3/2) - P(Xleqslant 1/2)$$
By definition, the last term is $P(Xleqslant 1/2)=F(1/2)$.
In other words, being strictly between $1/2$ and $3/2$ (that is, $1/2<X<3/2$) means being smaller than $3/2$ (that is, $X<3/2$) but not smaller than or equal to $1/2$ (that is, $X leqslant 1/2$).
If $X=1/2$, then $X$ does not satisfy $1/2 < X < 3/2$.
answered 4 hours ago
Taladris
4,46931732
answered 4 hours ago
Taladris
4,46931732
answered 4 hours ago
Taladris
4,46931732
answered 4 hours ago
Taladris
4,46931732
4,46931732
add a comment |Â
add a comment |Â
up vote
3
down vote
Your professor's answer is correct
Remember that you are dealing with a discrete random variable, so you can't just assume that the probability of being exactly $1/2$ is zero, so you need to remember to subtract it off.
Your solution finds $P(1/2 le X lt 3/2)$ which isnt necessarily the same for discrete random variables
answered 4 hours ago
WaveX
2,1552720
add a comment |Â
up vote
3
down vote
Your professor's answer is correct
Remember that you are dealing with a discrete random variable, so you can't just assume that the probability of being exactly $1/2$ is zero, so you need to remember to subtract it off.
Your solution finds $P(1/2 le X lt 3/2)$ which isnt necessarily the same for discrete random variables
answered 4 hours ago
WaveX
2,1552720
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Your professor's answer is correct
Remember that you are dealing with a discrete random variable, so you can't just assume that the probability of being exactly $1/2$ is zero, so you need to remember to subtract it off.
Your solution finds $P(1/2 le X lt 3/2)$ which isnt necessarily the same for discrete random variables
answered 4 hours ago
WaveX
2,1552720
Your professor's answer is correct
Remember that you are dealing with a discrete random variable, so you can't just assume that the probability of being exactly $1/2$ is zero, so you need to remember to subtract it off.
Your solution finds $P(1/2 le X lt 3/2)$ which isnt necessarily the same for discrete random variables
answered 4 hours ago
WaveX
2,1552720
answered 4 hours ago
WaveX
2,1552720
answered 4 hours ago
WaveX
2,1552720
answered 4 hours ago
WaveX
2,1552720
2,1552720
add a comment |Â
add a comment |Â
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We have $$X<3/2 = Xleq 1/2 cup 1/2<X<3/2$$ where the right-hand-side is a union of disjoint events, and so $$P[X<3/2] = P[Xleq 1/2] + P[1/2<X<3/2]$$
– Michael
4 hours ago
Note that $P[a < X le b] = F(b)-F(a)$.
– copper.hat
4 hours ago