Mixing
substitution in order to find the probability density function
$begingroup$
I was wondering how do you find the density function of a random variable $Y$ given that :
$X$ is an arbitrary random variable.
$Y := X^2$
I know how to do that for the case where $X$ is continuous. But how would you that in the discrete case?
In the continuous case you would search for $ P(Y leq y ) = P( X^2 leq y ) $, and then when you found the distributition function, you just have to derivate it. But without this precious tool, what are the ways to conclude ?
probability probability-theory probability-distributions
asked Jan 28 at 13:00
Marine GalantinMarine Galantin
940419
$endgroup$
add a comment |
$begingroup$
I was wondering how do you find the density function of a random variable $Y$ given that :
$X$ is an arbitrary random variable.
$Y := X^2$
I know how to do that for the case where $X$ is continuous. But how would you that in the discrete case?
In the continuous case you would search for $ P(Y leq y ) = P( X^2 leq y ) $, and then when you found the distributition function, you just have to derivate it. But without this precious tool, what are the ways to conclude ?
probability probability-theory probability-distributions
asked Jan 28 at 13:00
Marine GalantinMarine Galantin
940419
$endgroup$
$begingroup$
But in the discrete case you just have $P(Y =y) = P(X^2 = y)$. You don't need to use the distribution function $P(Y leq y)$.
$endgroup$
– Alex
Jan 28 at 13:10
1
$begingroup$
so the result is simply given by $P(Y = y) = P(X^2 = y) = P(X = +/- sqrt y) $ ?
$endgroup$
– Marine Galantin
Jan 28 at 13:17
1
$begingroup$
It's even easier in the discrete case. For example, if you want to find the probability of $P(Y=a)$, you know that it's equal to $P(Y=a)=P(X=-sqrt{a})+P(X=+sqrt{a})$
$endgroup$
– stressed out
Jan 28 at 13:17
add a comment |
$begingroup$
I was wondering how do you find the density function of a random variable $Y$ given that :
$X$ is an arbitrary random variable.
$Y := X^2$
I know how to do that for the case where $X$ is continuous. But how would you that in the discrete case?
In the continuous case you would search for $ P(Y leq y ) = P( X^2 leq y ) $, and then when you found the distributition function, you just have to derivate it. But without this precious tool, what are the ways to conclude ?
probability probability-theory probability-distributions
asked Jan 28 at 13:00
Marine GalantinMarine Galantin
940419
$endgroup$
I was wondering how do you find the density function of a random variable $Y$ given that :
$X$ is an arbitrary random variable.
$Y := X^2$
I know how to do that for the case where $X$ is continuous. But how would you that in the discrete case?
In the continuous case you would search for $ P(Y leq y ) = P( X^2 leq y ) $, and then when you found the distributition function, you just have to derivate it. But without this precious tool, what are the ways to conclude ?
probability probability-theory probability-distributions
probability probability-theory probability-distributions
asked Jan 28 at 13:00
Marine GalantinMarine Galantin
940419
asked Jan 28 at 13:00
Marine GalantinMarine Galantin
940419
asked Jan 28 at 13:00
Marine GalantinMarine Galantin
940419
asked Jan 28 at 13:00
Marine GalantinMarine Galantin
940419
asked Jan 28 at 13:00
Marine GalantinMarine Galantin
940419
940419
$begingroup$
But in the discrete case you just have $P(Y =y) = P(X^2 = y)$. You don't need to use the distribution function $P(Y leq y)$.
$endgroup$
– Alex
Jan 28 at 13:10
1
$begingroup$
so the result is simply given by $P(Y = y) = P(X^2 = y) = P(X = +/- sqrt y) $ ?
$endgroup$
– Marine Galantin
Jan 28 at 13:17
1
$begingroup$
It's even easier in the discrete case. For example, if you want to find the probability of $P(Y=a)$, you know that it's equal to $P(Y=a)=P(X=-sqrt{a})+P(X=+sqrt{a})$
$endgroup$
– stressed out
Jan 28 at 13:17
add a comment |
$begingroup$
But in the discrete case you just have $P(Y =y) = P(X^2 = y)$. You don't need to use the distribution function $P(Y leq y)$.
$endgroup$
– Alex
Jan 28 at 13:10
1
$begingroup$
so the result is simply given by $P(Y = y) = P(X^2 = y) = P(X = +/- sqrt y) $ ?
$endgroup$
– Marine Galantin
Jan 28 at 13:17
1
$begingroup$
It's even easier in the discrete case. For example, if you want to find the probability of $P(Y=a)$, you know that it's equal to $P(Y=a)=P(X=-sqrt{a})+P(X=+sqrt{a})$
$endgroup$
– stressed out
Jan 28 at 13:17
$begingroup$
But in the discrete case you just have $P(Y =y) = P(X^2 = y)$. You don't need to use the distribution function $P(Y leq y)$.
$endgroup$
– Alex
Jan 28 at 13:10
$begingroup$
But in the discrete case you just have $P(Y =y) = P(X^2 = y)$. You don't need to use the distribution function $P(Y leq y)$.
$endgroup$
– Alex
Jan 28 at 13:10
1
1
$begingroup$
so the result is simply given by $P(Y = y) = P(X^2 = y) = P(X = +/- sqrt y) $ ?
$endgroup$
– Marine Galantin
Jan 28 at 13:17
$begingroup$
so the result is simply given by $P(Y = y) = P(X^2 = y) = P(X = +/- sqrt y) $ ?
$endgroup$
– Marine Galantin
Jan 28 at 13:17
1
1
$begingroup$
It's even easier in the discrete case. For example, if you want to find the probability of $P(Y=a)$, you know that it's equal to $P(Y=a)=P(X=-sqrt{a})+P(X=+sqrt{a})$
$endgroup$
– stressed out
Jan 28 at 13:17
$begingroup$
It's even easier in the discrete case. For example, if you want to find the probability of $P(Y=a)$, you know that it's equal to $P(Y=a)=P(X=-sqrt{a})+P(X=+sqrt{a})$
$endgroup$
– stressed out
Jan 28 at 13:17
add a comment |
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$begingroup$
But in the discrete case you just have $P(Y =y) = P(X^2 = y)$. You don't need to use the distribution function $P(Y leq y)$.
$endgroup$
– Alex
Jan 28 at 13:10
1
$begingroup$
so the result is simply given by $P(Y = y) = P(X^2 = y) = P(X = +/- sqrt y) $ ?
$endgroup$
– Marine Galantin
Jan 28 at 13:17
1
$begingroup$
It's even easier in the discrete case. For example, if you want to find the probability of $P(Y=a)$, you know that it's equal to $P(Y=a)=P(X=-sqrt{a})+P(X=+sqrt{a})$
$endgroup$
– stressed out
Jan 28 at 13:17