Conditional probability computation
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I am having a hard time understanding something that supposedly should be very simple. I have a random variable $X$, taking values in the set of integers ${0,1,ldots,11}$ and is uniformly distributed in this set.
I also have a function $f(X)$ that given a realization of $X$, $x$ gives in output the binary representation of $x$ if $0leq x leq 7$ of $x-8$ otherwise.
Then it says that "clearly"
$$ P(x:f(x)=y| |y|=3)= frac{1/12}{8/12} = 1/8$$
Of course, the $P(|y|=3)= 8/12$ as it corresponds to the probability of $xin{0,1,ldots,7}$. But the numerator confuses me: why is $P(i:f(x)=y land|y|=3)$ still $1/12$?
Thank you for any suggestion you will provide me.
probability random-variables
asked yesterday
user1868607
1991110
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up vote
1
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favorite
I am having a hard time understanding something that supposedly should be very simple. I have a random variable $X$, taking values in the set of integers ${0,1,ldots,11}$ and is uniformly distributed in this set.
I also have a function $f(X)$ that given a realization of $X$, $x$ gives in output the binary representation of $x$ if $0leq x leq 7$ of $x-8$ otherwise.
Then it says that "clearly"
$$ P(x:f(x)=y| |y|=3)= frac{1/12}{8/12} = 1/8$$
Of course, the $P(|y|=3)= 8/12$ as it corresponds to the probability of $xin{0,1,ldots,7}$. But the numerator confuses me: why is $P(i:f(x)=y land|y|=3)$ still $1/12$?
Thank you for any suggestion you will provide me.
probability random-variables
asked yesterday
user1868607
1991110
Is there a typo at $...f(x)=y||y|=3...$? And how does your $f(x)$ look like?
– callculus
yesterday
Not a typo, it's the notation used in the notes : probability of y=f(x) conditioned on the length of y being fixed to 3. The function is described above: given a value x it outputs the binary representation in three bits if x is smaller or equal to 7, in two bits of x-7 if x is greater than 7.
– user1868607
yesterday
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am having a hard time understanding something that supposedly should be very simple. I have a random variable $X$, taking values in the set of integers ${0,1,ldots,11}$ and is uniformly distributed in this set.
I also have a function $f(X)$ that given a realization of $X$, $x$ gives in output the binary representation of $x$ if $0leq x leq 7$ of $x-8$ otherwise.
Then it says that "clearly"
$$ P(x:f(x)=y| |y|=3)= frac{1/12}{8/12} = 1/8$$
Of course, the $P(|y|=3)= 8/12$ as it corresponds to the probability of $xin{0,1,ldots,7}$. But the numerator confuses me: why is $P(i:f(x)=y land|y|=3)$ still $1/12$?
Thank you for any suggestion you will provide me.
probability random-variables
asked yesterday
user1868607
1991110
I am having a hard time understanding something that supposedly should be very simple. I have a random variable $X$, taking values in the set of integers ${0,1,ldots,11}$ and is uniformly distributed in this set.
I also have a function $f(X)$ that given a realization of $X$, $x$ gives in output the binary representation of $x$ if $0leq x leq 7$ of $x-8$ otherwise.
Then it says that "clearly"
$$ P(x:f(x)=y| |y|=3)= frac{1/12}{8/12} = 1/8$$
Of course, the $P(|y|=3)= 8/12$ as it corresponds to the probability of $xin{0,1,ldots,7}$. But the numerator confuses me: why is $P(i:f(x)=y land|y|=3)$ still $1/12$?
Thank you for any suggestion you will provide me.
probability random-variables
probability random-variables
asked yesterday
user1868607
1991110
asked yesterday
user1868607
1991110
edited yesterday
asked yesterday
user1868607
1991110
asked yesterday
user1868607
1991110
asked yesterday
user1868607
1991110
1991110
Is there a typo at $...f(x)=y||y|=3...$? And how does your $f(x)$ look like?
– callculus
yesterday
Not a typo, it's the notation used in the notes : probability of y=f(x) conditioned on the length of y being fixed to 3. The function is described above: given a value x it outputs the binary representation in three bits if x is smaller or equal to 7, in two bits of x-7 if x is greater than 7.
– user1868607
yesterday
add a comment |
Is there a typo at $...f(x)=y||y|=3...$? And how does your $f(x)$ look like?
– callculus
yesterday
Not a typo, it's the notation used in the notes : probability of y=f(x) conditioned on the length of y being fixed to 3. The function is described above: given a value x it outputs the binary representation in three bits if x is smaller or equal to 7, in two bits of x-7 if x is greater than 7.
– user1868607
yesterday
Is there a typo at $...f(x)=y||y|=3...$? And how does your $f(x)$ look like?
– callculus
yesterday
Is there a typo at $...f(x)=y||y|=3...$? And how does your $f(x)$ look like?
– callculus
yesterday
Not a typo, it's the notation used in the notes : probability of y=f(x) conditioned on the length of y being fixed to 3. The function is described above: given a value x it outputs the binary representation in three bits if x is smaller or equal to 7, in two bits of x-7 if x is greater than 7.
– user1868607
yesterday
Not a typo, it's the notation used in the notes : probability of y=f(x) conditioned on the length of y being fixed to 3. The function is described above: given a value x it outputs the binary representation in three bits if x is smaller or equal to 7, in two bits of x-7 if x is greater than 7.
– user1868607
yesterday
add a comment |
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Is there a typo at $...f(x)=y||y|=3...$? And how does your $f(x)$ look like?
– callculus
yesterday
Not a typo, it's the notation used in the notes : probability of y=f(x) conditioned on the length of y being fixed to 3. The function is described above: given a value x it outputs the binary representation in three bits if x is smaller or equal to 7, in two bits of x-7 if x is greater than 7.
– user1868607
yesterday