How to determine cumulative distribution function of squared random variable?
$begingroup$
I don't understand how can I get the
$F(x)$(cumulative distribution funciton)
given only
$X = RND^2$
where RND means (continuous) random variable, the paper tells the answer is $F(x) = sqrt x$
, if x $epsilon (0,1)$ it seems that I need to take square root but why?
I know $F(x) = p(X < x)$ but from here I can't proceed.
probability probability-theory probability-distributions random-variables
asked Dec 13 '18 at 17:26
Levente BartosLevente Bartos
186
$endgroup$
add a comment |
$begingroup$
I don't understand how can I get the
$F(x)$(cumulative distribution funciton)
given only
$X = RND^2$
where RND means (continuous) random variable, the paper tells the answer is $F(x) = sqrt x$
, if x $epsilon (0,1)$ it seems that I need to take square root but why?
I know $F(x) = p(X < x)$ but from here I can't proceed.
probability probability-theory probability-distributions random-variables
asked Dec 13 '18 at 17:26
Levente BartosLevente Bartos
186
$endgroup$
$begingroup$
Do you know the distribution function for RND?
$endgroup$
– herb steinberg
Dec 13 '18 at 17:30
$begingroup$
I have added a picture about the problem, i assume if I get this kind of question during an exam they will give only the first line X = RND^2 and then i should continue?
$endgroup$
– Levente Bartos
Dec 13 '18 at 17:40
$begingroup$
I assume it is uniformly distributed
$endgroup$
– Levente Bartos
Dec 13 '18 at 17:42
add a comment |
$begingroup$
I don't understand how can I get the
$F(x)$(cumulative distribution funciton)
given only
$X = RND^2$
where RND means (continuous) random variable, the paper tells the answer is $F(x) = sqrt x$
, if x $epsilon (0,1)$ it seems that I need to take square root but why?
I know $F(x) = p(X < x)$ but from here I can't proceed.
probability probability-theory probability-distributions random-variables
asked Dec 13 '18 at 17:26
Levente BartosLevente Bartos
186
$endgroup$
I don't understand how can I get the
$F(x)$(cumulative distribution funciton)
given only
$X = RND^2$
where RND means (continuous) random variable, the paper tells the answer is $F(x) = sqrt x$
, if x $epsilon (0,1)$ it seems that I need to take square root but why?
I know $F(x) = p(X < x)$ but from here I can't proceed.
probability probability-theory probability-distributions random-variables
probability probability-theory probability-distributions random-variables
asked Dec 13 '18 at 17:26
Levente BartosLevente Bartos
186
asked Dec 13 '18 at 17:26
Levente BartosLevente Bartos
186
edited Dec 13 '18 at 17:39
Levente Bartos
asked Dec 13 '18 at 17:26
Levente BartosLevente Bartos
186
asked Dec 13 '18 at 17:26
Levente BartosLevente Bartos
186
asked Dec 13 '18 at 17:26
Levente BartosLevente Bartos
186
186
$begingroup$
Do you know the distribution function for RND?
$endgroup$
– herb steinberg
Dec 13 '18 at 17:30
$begingroup$
I have added a picture about the problem, i assume if I get this kind of question during an exam they will give only the first line X = RND^2 and then i should continue?
$endgroup$
– Levente Bartos
Dec 13 '18 at 17:40
$begingroup$
I assume it is uniformly distributed
$endgroup$
– Levente Bartos
Dec 13 '18 at 17:42
add a comment |
$begingroup$
Do you know the distribution function for RND?
$endgroup$
– herb steinberg
Dec 13 '18 at 17:30
$begingroup$
I have added a picture about the problem, i assume if I get this kind of question during an exam they will give only the first line X = RND^2 and then i should continue?
$endgroup$
– Levente Bartos
Dec 13 '18 at 17:40
$begingroup$
I assume it is uniformly distributed
$endgroup$
– Levente Bartos
Dec 13 '18 at 17:42
$begingroup$
Do you know the distribution function for RND?
$endgroup$
– herb steinberg
Dec 13 '18 at 17:30
$begingroup$
Do you know the distribution function for RND?
$endgroup$
– herb steinberg
Dec 13 '18 at 17:30
$begingroup$
I have added a picture about the problem, i assume if I get this kind of question during an exam they will give only the first line X = RND^2 and then i should continue?
$endgroup$
– Levente Bartos
Dec 13 '18 at 17:40
$begingroup$
I have added a picture about the problem, i assume if I get this kind of question during an exam they will give only the first line X = RND^2 and then i should continue?
$endgroup$
– Levente Bartos
Dec 13 '18 at 17:40
$begingroup$
I assume it is uniformly distributed
$endgroup$
– Levente Bartos
Dec 13 '18 at 17:42
$begingroup$
I assume it is uniformly distributed
$endgroup$
– Levente Bartos
Dec 13 '18 at 17:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have $X=Y^2$, where $Y sim U(0,1)$. Now we want the cdf of $X$.
$P(X<x)=P(Y^2<x)=P(Y<sqrt x)=F_Y(sqrt x)$
Now you plug in the value $sqrt x$ into the cdf of $Y$:
$$F_Y(y)=begin{cases}
0 & text{for } y < 0 \
y & text{for } y in [0,1) \
1 & text{for } x ge 1
end{cases}$$
To obtain the pdf you just differentiate the cdf of Y.
answered Dec 13 '18 at 17:48
callculuscallculus
18.2k31427
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have $X=Y^2$, where $Y sim U(0,1)$. Now we want the cdf of $X$.
$P(X<x)=P(Y^2<x)=P(Y<sqrt x)=F_Y(sqrt x)$
Now you plug in the value $sqrt x$ into the cdf of $Y$:
$$F_Y(y)=begin{cases}
0 & text{for } y < 0 \
y & text{for } y in [0,1) \
1 & text{for } x ge 1
end{cases}$$
To obtain the pdf you just differentiate the cdf of Y.
answered Dec 13 '18 at 17:48
callculuscallculus
18.2k31427
$endgroup$
add a comment |
$begingroup$
You have $X=Y^2$, where $Y sim U(0,1)$. Now we want the cdf of $X$.
$P(X<x)=P(Y^2<x)=P(Y<sqrt x)=F_Y(sqrt x)$
Now you plug in the value $sqrt x$ into the cdf of $Y$:
$$F_Y(y)=begin{cases}
0 & text{for } y < 0 \
y & text{for } y in [0,1) \
1 & text{for } x ge 1
end{cases}$$
To obtain the pdf you just differentiate the cdf of Y.
answered Dec 13 '18 at 17:48
callculuscallculus
18.2k31427
$endgroup$
add a comment |
$begingroup$
You have $X=Y^2$, where $Y sim U(0,1)$. Now we want the cdf of $X$.
$P(X<x)=P(Y^2<x)=P(Y<sqrt x)=F_Y(sqrt x)$
Now you plug in the value $sqrt x$ into the cdf of $Y$:
$$F_Y(y)=begin{cases}
0 & text{for } y < 0 \
y & text{for } y in [0,1) \
1 & text{for } x ge 1
end{cases}$$
To obtain the pdf you just differentiate the cdf of Y.
answered Dec 13 '18 at 17:48
callculuscallculus
18.2k31427
$endgroup$
You have $X=Y^2$, where $Y sim U(0,1)$. Now we want the cdf of $X$.
$P(X<x)=P(Y^2<x)=P(Y<sqrt x)=F_Y(sqrt x)$
Now you plug in the value $sqrt x$ into the cdf of $Y$:
$$F_Y(y)=begin{cases}
0 & text{for } y < 0 \
y & text{for } y in [0,1) \
1 & text{for } x ge 1
end{cases}$$
To obtain the pdf you just differentiate the cdf of Y.
answered Dec 13 '18 at 17:48
callculuscallculus
18.2k31427
answered Dec 13 '18 at 17:48
callculuscallculus
18.2k31427
answered Dec 13 '18 at 17:48
callculuscallculus
18.2k31427
answered Dec 13 '18 at 17:48
callculuscallculus
18.2k31427
18.2k31427
add a comment |
add a comment |
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$begingroup$
Do you know the distribution function for RND?
$endgroup$
– herb steinberg
Dec 13 '18 at 17:30
$begingroup$
I have added a picture about the problem, i assume if I get this kind of question during an exam they will give only the first line X = RND^2 and then i should continue?
$endgroup$
– Levente Bartos
Dec 13 '18 at 17:40
$begingroup$
I assume it is uniformly distributed
$endgroup$
– Levente Bartos
Dec 13 '18 at 17:42