If $Xsim Unif[0,1]$ what is $f_Y$ if $Y=g(X)$?
Let $Xsim Uniform[0,1]$ and $$g(x)=begin{cases}3x&xin [0,1/3]\ 0&otherwiseend{cases}.$$
What is the density function of $Y$ ? (denoted $f_Y$).
I'm a bit confuse. The cumulative function is given by $$F_Y(y)=mathbb P{Yleq y}=mathbb P{g(X)leq y},$$
but I have problem to continue. Is it $$mathbb P{G(X)leq y}=mathbb P{G(X)leq y, Xin [0,1]}$$
$$=mathbb P{g(X)leq y,Xin[0,1/3]}+mathbb P{g(X)leq y, Xin [1/3,1]}=mathbb P{3Xleq y, Xin [0,1/3]}+mathbb P{0leq y,Xin [1/3,1]} ?$$
I'm confuse on how to continue.
probability
asked Nov 29 '18 at 20:19
user621128user621128
161
add a comment |
Let $Xsim Uniform[0,1]$ and $$g(x)=begin{cases}3x&xin [0,1/3]\ 0&otherwiseend{cases}.$$
What is the density function of $Y$ ? (denoted $f_Y$).
I'm a bit confuse. The cumulative function is given by $$F_Y(y)=mathbb P{Yleq y}=mathbb P{g(X)leq y},$$
but I have problem to continue. Is it $$mathbb P{G(X)leq y}=mathbb P{G(X)leq y, Xin [0,1]}$$
$$=mathbb P{g(X)leq y,Xin[0,1/3]}+mathbb P{g(X)leq y, Xin [1/3,1]}=mathbb P{3Xleq y, Xin [0,1/3]}+mathbb P{0leq y,Xin [1/3,1]} ?$$
I'm confuse on how to continue.
probability
asked Nov 29 '18 at 20:19
user621128user621128
161
add a comment |
Let $Xsim Uniform[0,1]$ and $$g(x)=begin{cases}3x&xin [0,1/3]\ 0&otherwiseend{cases}.$$
What is the density function of $Y$ ? (denoted $f_Y$).
I'm a bit confuse. The cumulative function is given by $$F_Y(y)=mathbb P{Yleq y}=mathbb P{g(X)leq y},$$
but I have problem to continue. Is it $$mathbb P{G(X)leq y}=mathbb P{G(X)leq y, Xin [0,1]}$$
$$=mathbb P{g(X)leq y,Xin[0,1/3]}+mathbb P{g(X)leq y, Xin [1/3,1]}=mathbb P{3Xleq y, Xin [0,1/3]}+mathbb P{0leq y,Xin [1/3,1]} ?$$
I'm confuse on how to continue.
probability
asked Nov 29 '18 at 20:19
user621128user621128
161
Let $Xsim Uniform[0,1]$ and $$g(x)=begin{cases}3x&xin [0,1/3]\ 0&otherwiseend{cases}.$$
What is the density function of $Y$ ? (denoted $f_Y$).
I'm a bit confuse. The cumulative function is given by $$F_Y(y)=mathbb P{Yleq y}=mathbb P{g(X)leq y},$$
but I have problem to continue. Is it $$mathbb P{G(X)leq y}=mathbb P{G(X)leq y, Xin [0,1]}$$
$$=mathbb P{g(X)leq y,Xin[0,1/3]}+mathbb P{g(X)leq y, Xin [1/3,1]}=mathbb P{3Xleq y, Xin [0,1/3]}+mathbb P{0leq y,Xin [1/3,1]} ?$$
I'm confuse on how to continue.
probability
probability
asked Nov 29 '18 at 20:19
user621128user621128
161
asked Nov 29 '18 at 20:19
user621128user621128
161
asked Nov 29 '18 at 20:19
user621128user621128
161
asked Nov 29 '18 at 20:19
user621128user621128
161
asked Nov 29 '18 at 20:19
user621128user621128
161
161
add a comment |
add a comment |
2 Answers
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The truth is $Y$ has no density function, since it's not a continuous random variable. It's actually not a discrete one, either.
Note that $P(Y=0)=Pleft(Xnotin left[0,frac13right]right)=frac23$, so $Y$ is not continuous, since it has positive probability in one point ($Y=0$), but this is the only point with non-zero probability, so that $Y$ can't be discrete (because only $frac23$ of the probability are accumulated at points, the other $frac13$ is continuously accumulated in the interval $(0,1)$.
That is, $Y$ has a mixed type probability distribution.
While it has no PDF nor PMF, it has a CDF (any random variable does) and it is
$$F_Y(y)=left{begin{matrix}0 & y<0\ frac23+frac y3 & 0le y<1\ 1 & yge 1.end{matrix}right.$$
answered Nov 29 '18 at 20:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,409118
Nice answer. I was in the wrong way since the beginning. (+1)
– Surb
Nov 29 '18 at 21:11
But for $yin (0,1)$ can't we say that $f_Y(y)=frac{1}{3}$ ? (i.e. it has a density for all $yin (0,1)$ ?
– Surb
Nov 29 '18 at 21:32
Well... that would not be a density since it does not integrate $1$. We could say that $Y$ has a density conditioned on $Yneq 0$, though, but that density would be $1$, not $frac 13$. If $Yneq 0$, then $Ysim mathbb U(0,1)$, but if $Y=0$, then... well... then $Y=0$, and that's it.
– Alejandro Nasif Salum
Nov 29 '18 at 21:46
Another way to see a mixed r.v. is to take a r.v. $U$ which is $0$ with probability $frac13$ and $1$ with probability $frac23$ (that is, a Bernoulli r.v. with $p=frac23$). Then define: $$Y=left{begin{matrix}W& U=0\ 0 & U=1\end{matrix}right.$$
– Alejandro Nasif Salum
Nov 29 '18 at 21:48
add a comment |
A simpler approach is
$$
f_X{rm d}X = f_Y {rm d}Y ~~Rightarrow ~~~ f_Y(y) = f_X(x)left|frac{{rm d}X}{{rm d}Y}right| = 1 cdot frac{1}{3} = frac{1}{3}
$$
answered Nov 29 '18 at 20:32
caveraccaverac
14k21130
add a comment |
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2 Answers
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2 Answers
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The truth is $Y$ has no density function, since it's not a continuous random variable. It's actually not a discrete one, either.
Note that $P(Y=0)=Pleft(Xnotin left[0,frac13right]right)=frac23$, so $Y$ is not continuous, since it has positive probability in one point ($Y=0$), but this is the only point with non-zero probability, so that $Y$ can't be discrete (because only $frac23$ of the probability are accumulated at points, the other $frac13$ is continuously accumulated in the interval $(0,1)$.
That is, $Y$ has a mixed type probability distribution.
While it has no PDF nor PMF, it has a CDF (any random variable does) and it is
$$F_Y(y)=left{begin{matrix}0 & y<0\ frac23+frac y3 & 0le y<1\ 1 & yge 1.end{matrix}right.$$
answered Nov 29 '18 at 20:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,409118
Nice answer. I was in the wrong way since the beginning. (+1)
– Surb
Nov 29 '18 at 21:11
But for $yin (0,1)$ can't we say that $f_Y(y)=frac{1}{3}$ ? (i.e. it has a density for all $yin (0,1)$ ?
– Surb
Nov 29 '18 at 21:32
Well... that would not be a density since it does not integrate $1$. We could say that $Y$ has a density conditioned on $Yneq 0$, though, but that density would be $1$, not $frac 13$. If $Yneq 0$, then $Ysim mathbb U(0,1)$, but if $Y=0$, then... well... then $Y=0$, and that's it.
– Alejandro Nasif Salum
Nov 29 '18 at 21:46
Another way to see a mixed r.v. is to take a r.v. $U$ which is $0$ with probability $frac13$ and $1$ with probability $frac23$ (that is, a Bernoulli r.v. with $p=frac23$). Then define: $$Y=left{begin{matrix}W& U=0\ 0 & U=1\end{matrix}right.$$
– Alejandro Nasif Salum
Nov 29 '18 at 21:48
add a comment |
The truth is $Y$ has no density function, since it's not a continuous random variable. It's actually not a discrete one, either.
Note that $P(Y=0)=Pleft(Xnotin left[0,frac13right]right)=frac23$, so $Y$ is not continuous, since it has positive probability in one point ($Y=0$), but this is the only point with non-zero probability, so that $Y$ can't be discrete (because only $frac23$ of the probability are accumulated at points, the other $frac13$ is continuously accumulated in the interval $(0,1)$.
That is, $Y$ has a mixed type probability distribution.
While it has no PDF nor PMF, it has a CDF (any random variable does) and it is
$$F_Y(y)=left{begin{matrix}0 & y<0\ frac23+frac y3 & 0le y<1\ 1 & yge 1.end{matrix}right.$$
answered Nov 29 '18 at 20:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,409118
Nice answer. I was in the wrong way since the beginning. (+1)
– Surb
Nov 29 '18 at 21:11
But for $yin (0,1)$ can't we say that $f_Y(y)=frac{1}{3}$ ? (i.e. it has a density for all $yin (0,1)$ ?
– Surb
Nov 29 '18 at 21:32
Well... that would not be a density since it does not integrate $1$. We could say that $Y$ has a density conditioned on $Yneq 0$, though, but that density would be $1$, not $frac 13$. If $Yneq 0$, then $Ysim mathbb U(0,1)$, but if $Y=0$, then... well... then $Y=0$, and that's it.
– Alejandro Nasif Salum
Nov 29 '18 at 21:46
Another way to see a mixed r.v. is to take a r.v. $U$ which is $0$ with probability $frac13$ and $1$ with probability $frac23$ (that is, a Bernoulli r.v. with $p=frac23$). Then define: $$Y=left{begin{matrix}W& U=0\ 0 & U=1\end{matrix}right.$$
– Alejandro Nasif Salum
Nov 29 '18 at 21:48
add a comment |
The truth is $Y$ has no density function, since it's not a continuous random variable. It's actually not a discrete one, either.
Note that $P(Y=0)=Pleft(Xnotin left[0,frac13right]right)=frac23$, so $Y$ is not continuous, since it has positive probability in one point ($Y=0$), but this is the only point with non-zero probability, so that $Y$ can't be discrete (because only $frac23$ of the probability are accumulated at points, the other $frac13$ is continuously accumulated in the interval $(0,1)$.
That is, $Y$ has a mixed type probability distribution.
While it has no PDF nor PMF, it has a CDF (any random variable does) and it is
$$F_Y(y)=left{begin{matrix}0 & y<0\ frac23+frac y3 & 0le y<1\ 1 & yge 1.end{matrix}right.$$
answered Nov 29 '18 at 20:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,409118
The truth is $Y$ has no density function, since it's not a continuous random variable. It's actually not a discrete one, either.
Note that $P(Y=0)=Pleft(Xnotin left[0,frac13right]right)=frac23$, so $Y$ is not continuous, since it has positive probability in one point ($Y=0$), but this is the only point with non-zero probability, so that $Y$ can't be discrete (because only $frac23$ of the probability are accumulated at points, the other $frac13$ is continuously accumulated in the interval $(0,1)$.
That is, $Y$ has a mixed type probability distribution.
While it has no PDF nor PMF, it has a CDF (any random variable does) and it is
$$F_Y(y)=left{begin{matrix}0 & y<0\ frac23+frac y3 & 0le y<1\ 1 & yge 1.end{matrix}right.$$
answered Nov 29 '18 at 20:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,409118
edited Nov 29 '18 at 21:03
answered Nov 29 '18 at 20:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,409118
answered Nov 29 '18 at 20:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,409118
answered Nov 29 '18 at 20:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,409118
4,409118
Nice answer. I was in the wrong way since the beginning. (+1)
– Surb
Nov 29 '18 at 21:11
But for $yin (0,1)$ can't we say that $f_Y(y)=frac{1}{3}$ ? (i.e. it has a density for all $yin (0,1)$ ?
– Surb
Nov 29 '18 at 21:32
Well... that would not be a density since it does not integrate $1$. We could say that $Y$ has a density conditioned on $Yneq 0$, though, but that density would be $1$, not $frac 13$. If $Yneq 0$, then $Ysim mathbb U(0,1)$, but if $Y=0$, then... well... then $Y=0$, and that's it.
– Alejandro Nasif Salum
Nov 29 '18 at 21:46
Another way to see a mixed r.v. is to take a r.v. $U$ which is $0$ with probability $frac13$ and $1$ with probability $frac23$ (that is, a Bernoulli r.v. with $p=frac23$). Then define: $$Y=left{begin{matrix}W& U=0\ 0 & U=1\end{matrix}right.$$
– Alejandro Nasif Salum
Nov 29 '18 at 21:48
add a comment |
Nice answer. I was in the wrong way since the beginning. (+1)
– Surb
Nov 29 '18 at 21:11
But for $yin (0,1)$ can't we say that $f_Y(y)=frac{1}{3}$ ? (i.e. it has a density for all $yin (0,1)$ ?
– Surb
Nov 29 '18 at 21:32
Well... that would not be a density since it does not integrate $1$. We could say that $Y$ has a density conditioned on $Yneq 0$, though, but that density would be $1$, not $frac 13$. If $Yneq 0$, then $Ysim mathbb U(0,1)$, but if $Y=0$, then... well... then $Y=0$, and that's it.
– Alejandro Nasif Salum
Nov 29 '18 at 21:46
Another way to see a mixed r.v. is to take a r.v. $U$ which is $0$ with probability $frac13$ and $1$ with probability $frac23$ (that is, a Bernoulli r.v. with $p=frac23$). Then define: $$Y=left{begin{matrix}W& U=0\ 0 & U=1\end{matrix}right.$$
– Alejandro Nasif Salum
Nov 29 '18 at 21:48
Nice answer. I was in the wrong way since the beginning. (+1)
– Surb
Nov 29 '18 at 21:11
Nice answer. I was in the wrong way since the beginning. (+1)
– Surb
Nov 29 '18 at 21:11
But for $yin (0,1)$ can't we say that $f_Y(y)=frac{1}{3}$ ? (i.e. it has a density for all $yin (0,1)$ ?
– Surb
Nov 29 '18 at 21:32
But for $yin (0,1)$ can't we say that $f_Y(y)=frac{1}{3}$ ? (i.e. it has a density for all $yin (0,1)$ ?
– Surb
Nov 29 '18 at 21:32
Well... that would not be a density since it does not integrate $1$. We could say that $Y$ has a density conditioned on $Yneq 0$, though, but that density would be $1$, not $frac 13$. If $Yneq 0$, then $Ysim mathbb U(0,1)$, but if $Y=0$, then... well... then $Y=0$, and that's it.
– Alejandro Nasif Salum
Nov 29 '18 at 21:46
Well... that would not be a density since it does not integrate $1$. We could say that $Y$ has a density conditioned on $Yneq 0$, though, but that density would be $1$, not $frac 13$. If $Yneq 0$, then $Ysim mathbb U(0,1)$, but if $Y=0$, then... well... then $Y=0$, and that's it.
– Alejandro Nasif Salum
Nov 29 '18 at 21:46
Another way to see a mixed r.v. is to take a r.v. $U$ which is $0$ with probability $frac13$ and $1$ with probability $frac23$ (that is, a Bernoulli r.v. with $p=frac23$). Then define: $$Y=left{begin{matrix}W& U=0\ 0 & U=1\end{matrix}right.$$
– Alejandro Nasif Salum
Nov 29 '18 at 21:48
Another way to see a mixed r.v. is to take a r.v. $U$ which is $0$ with probability $frac13$ and $1$ with probability $frac23$ (that is, a Bernoulli r.v. with $p=frac23$). Then define: $$Y=left{begin{matrix}W& U=0\ 0 & U=1\end{matrix}right.$$
– Alejandro Nasif Salum
Nov 29 '18 at 21:48
add a comment |
A simpler approach is
$$
f_X{rm d}X = f_Y {rm d}Y ~~Rightarrow ~~~ f_Y(y) = f_X(x)left|frac{{rm d}X}{{rm d}Y}right| = 1 cdot frac{1}{3} = frac{1}{3}
$$
answered Nov 29 '18 at 20:32
caveraccaverac
14k21130
add a comment |
A simpler approach is
$$
f_X{rm d}X = f_Y {rm d}Y ~~Rightarrow ~~~ f_Y(y) = f_X(x)left|frac{{rm d}X}{{rm d}Y}right| = 1 cdot frac{1}{3} = frac{1}{3}
$$
answered Nov 29 '18 at 20:32
caveraccaverac
14k21130
add a comment |
A simpler approach is
$$
f_X{rm d}X = f_Y {rm d}Y ~~Rightarrow ~~~ f_Y(y) = f_X(x)left|frac{{rm d}X}{{rm d}Y}right| = 1 cdot frac{1}{3} = frac{1}{3}
$$
answered Nov 29 '18 at 20:32
caveraccaverac
14k21130
A simpler approach is
$$
f_X{rm d}X = f_Y {rm d}Y ~~Rightarrow ~~~ f_Y(y) = f_X(x)left|frac{{rm d}X}{{rm d}Y}right| = 1 cdot frac{1}{3} = frac{1}{3}
$$
answered Nov 29 '18 at 20:32
caveraccaverac
14k21130
edited Nov 29 '18 at 20:38
answered Nov 29 '18 at 20:32
caveraccaverac
14k21130
answered Nov 29 '18 at 20:32
caveraccaverac
14k21130
answered Nov 29 '18 at 20:32
caveraccaverac
14k21130
14k21130
add a comment |
add a comment |
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