Mixing
Calculate distribution of Y - can't figure out a specific range
$begingroup$
Let $ X sim U[-1,2]$ and
$$
Y =
begin{cases}
X^2 & X^2leq 2 \
2 & X^2>2
end{cases}
$$
I need to find the distribution function of Y.
So as usual, in order to find $ F_Y(t)$ i separated to cases depending on t values, according to the plot:
So my cases are: $ t<0 $, $0leq t < 2$ and $t>2$.
For some reason, the (not detailed) solution separated to the following cases: $t<0$, $0leq t <1$, $1leq t <2$ and $ t>2$. Why is that? is it related to the range of X?
Thanks!
probability probability-distributions
asked Jan 29 at 17:07
superuser123superuser123
48628
$endgroup$
add a comment |
$begingroup$
Let $ X sim U[-1,2]$ and
$$
Y =
begin{cases}
X^2 & X^2leq 2 \
2 & X^2>2
end{cases}
$$
I need to find the distribution function of Y.
So as usual, in order to find $ F_Y(t)$ i separated to cases depending on t values, according to the plot:
So my cases are: $ t<0 $, $0leq t < 2$ and $t>2$.
For some reason, the (not detailed) solution separated to the following cases: $t<0$, $0leq t <1$, $1leq t <2$ and $ t>2$. Why is that? is it related to the range of X?
Thanks!
probability probability-distributions
asked Jan 29 at 17:07
superuser123superuser123
48628
$endgroup$
$begingroup$
Note that you can get 0<t<1 for -1<x<1 and 1<t you can get only for 1<x. I would expect the rate of the function for t<1 to be double than 1<t<2
$endgroup$
– Shaq
Jan 30 at 9:54
add a comment |
$begingroup$
Let $ X sim U[-1,2]$ and
$$
Y =
begin{cases}
X^2 & X^2leq 2 \
2 & X^2>2
end{cases}
$$
I need to find the distribution function of Y.
So as usual, in order to find $ F_Y(t)$ i separated to cases depending on t values, according to the plot:
So my cases are: $ t<0 $, $0leq t < 2$ and $t>2$.
For some reason, the (not detailed) solution separated to the following cases: $t<0$, $0leq t <1$, $1leq t <2$ and $ t>2$. Why is that? is it related to the range of X?
Thanks!
probability probability-distributions
asked Jan 29 at 17:07
superuser123superuser123
48628
$endgroup$
Let $ X sim U[-1,2]$ and
$$
Y =
begin{cases}
X^2 & X^2leq 2 \
2 & X^2>2
end{cases}
$$
I need to find the distribution function of Y.
So as usual, in order to find $ F_Y(t)$ i separated to cases depending on t values, according to the plot:
So my cases are: $ t<0 $, $0leq t < 2$ and $t>2$.
For some reason, the (not detailed) solution separated to the following cases: $t<0$, $0leq t <1$, $1leq t <2$ and $ t>2$. Why is that? is it related to the range of X?
Thanks!
probability probability-distributions
probability probability-distributions
asked Jan 29 at 17:07
superuser123superuser123
48628
asked Jan 29 at 17:07
superuser123superuser123
48628
asked Jan 29 at 17:07
superuser123superuser123
48628
asked Jan 29 at 17:07
superuser123superuser123
48628
asked Jan 29 at 17:07
superuser123superuser123
48628
48628
$begingroup$
Note that you can get 0<t<1 for -1<x<1 and 1<t you can get only for 1<x. I would expect the rate of the function for t<1 to be double than 1<t<2
$endgroup$
– Shaq
Jan 30 at 9:54
add a comment |
$begingroup$
Note that you can get 0<t<1 for -1<x<1 and 1<t you can get only for 1<x. I would expect the rate of the function for t<1 to be double than 1<t<2
$endgroup$
– Shaq
Jan 30 at 9:54
$begingroup$
Note that you can get 0<t<1 for -1<x<1 and 1<t you can get only for 1<x. I would expect the rate of the function for t<1 to be double than 1<t<2
$endgroup$
– Shaq
Jan 30 at 9:54
$begingroup$
Note that you can get 0<t<1 for -1<x<1 and 1<t you can get only for 1<x. I would expect the rate of the function for t<1 to be double than 1<t<2
$endgroup$
– Shaq
Jan 30 at 9:54
add a comment |
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$begingroup$
Note that you can get 0<t<1 for -1<x<1 and 1<t you can get only for 1<x. I would expect the rate of the function for t<1 to be double than 1<t<2
$endgroup$
– Shaq
Jan 30 at 9:54