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Find the distribution of random variable using specific method.
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Let $X$ be the random variable with distribution $X in mathcalE(1) $, find the distributions of the following random variables:
a) $Y=X^2$
b) $Y=lfloor X rfloor$
$f_X(x)= e^-x, xgeq 0$
NOTE: The only method i am allowed to use here (not sure if there's a name for that method) is the method that consists of drawing a graph of function and then placing few characteristic lines on graph parallel to the $X$-axis, and then calculating the integral between the points where line and graph intersect
For example:
Now, integrating from $-sqrt2 $ to $sqrt2$ gives us the distribution function for $Y$.
I would like an intuitive explanation on why this works and how could i find density of $Y$ when i have this. Any help appreciated!
probability
asked Jul 25 at 13:25
cdummie
689412
 |Â
show 4 more comments
up vote
0
down vote
favorite
Let $X$ be the random variable with distribution $X in mathcalE(1) $, find the distributions of the following random variables:
a) $Y=X^2$
b) $Y=lfloor X rfloor$
$f_X(x)= e^-x, xgeq 0$
NOTE: The only method i am allowed to use here (not sure if there's a name for that method) is the method that consists of drawing a graph of function and then placing few characteristic lines on graph parallel to the $X$-axis, and then calculating the integral between the points where line and graph intersect
For example:
Now, integrating from $-sqrt2 $ to $sqrt2$ gives us the distribution function for $Y$.
I would like an intuitive explanation on why this works and how could i find density of $Y$ when i have this. Any help appreciated!
probability
asked Jul 25 at 13:25
cdummie
689412
2
This is the usual CDF method. If $Y=g(X)$, then distribution function of $Y$ is $$P(Yle y)=P(g(X)le y)=P(Xin A),,quad A=x:g(x)le y$$ Using graph, you are finding the region $A$.
– StubbornAtom
Jul 25 at 15:34
Oh, i see, thanks, but how could i find density function when i have this?
– cdummie
Jul 26 at 7:12
1
By differentiating the distribution function, you would get the density function, provided $Y$ is absolutely continuous.
– StubbornAtom
Jul 26 at 8:14
@StubbornAtom That works because differentiation is opposite of integration in a way, i think i got that, what about the second case b) , as far as i can see that is not continuous distribution, how am i supposed to act in that kind of situations, will it's distribution also be discrete?
– cdummie
Jul 27 at 18:53
1
No. $P(lfloor Xrfloor=k)=P(kle X<k+1)=cdots$
– StubbornAtom
Jul 28 at 15:13
 |Â
show 4 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X$ be the random variable with distribution $X in mathcalE(1) $, find the distributions of the following random variables:
a) $Y=X^2$
b) $Y=lfloor X rfloor$
$f_X(x)= e^-x, xgeq 0$
NOTE: The only method i am allowed to use here (not sure if there's a name for that method) is the method that consists of drawing a graph of function and then placing few characteristic lines on graph parallel to the $X$-axis, and then calculating the integral between the points where line and graph intersect
For example:
Now, integrating from $-sqrt2 $ to $sqrt2$ gives us the distribution function for $Y$.
I would like an intuitive explanation on why this works and how could i find density of $Y$ when i have this. Any help appreciated!
probability
asked Jul 25 at 13:25
cdummie
689412
Let $X$ be the random variable with distribution $X in mathcalE(1) $, find the distributions of the following random variables:
a) $Y=X^2$
b) $Y=lfloor X rfloor$
$f_X(x)= e^-x, xgeq 0$
NOTE: The only method i am allowed to use here (not sure if there's a name for that method) is the method that consists of drawing a graph of function and then placing few characteristic lines on graph parallel to the $X$-axis, and then calculating the integral between the points where line and graph intersect
For example:
Now, integrating from $-sqrt2 $ to $sqrt2$ gives us the distribution function for $Y$.
I would like an intuitive explanation on why this works and how could i find density of $Y$ when i have this. Any help appreciated!
probability
asked Jul 25 at 13:25
cdummie
689412
asked Jul 25 at 13:25
cdummie
689412
asked Jul 25 at 13:25
cdummie
689412
asked Jul 25 at 13:25
cdummie
689412
689412
2
This is the usual CDF method. If $Y=g(X)$, then distribution function of $Y$ is $$P(Yle y)=P(g(X)le y)=P(Xin A),,quad A=x:g(x)le y$$ Using graph, you are finding the region $A$.
– StubbornAtom
Jul 25 at 15:34
Oh, i see, thanks, but how could i find density function when i have this?
– cdummie
Jul 26 at 7:12
1
By differentiating the distribution function, you would get the density function, provided $Y$ is absolutely continuous.
– StubbornAtom
Jul 26 at 8:14
@StubbornAtom That works because differentiation is opposite of integration in a way, i think i got that, what about the second case b) , as far as i can see that is not continuous distribution, how am i supposed to act in that kind of situations, will it's distribution also be discrete?
– cdummie
Jul 27 at 18:53
1
No. $P(lfloor Xrfloor=k)=P(kle X<k+1)=cdots$
– StubbornAtom
Jul 28 at 15:13
 |Â
show 4 more comments
2
This is the usual CDF method. If $Y=g(X)$, then distribution function of $Y$ is $$P(Yle y)=P(g(X)le y)=P(Xin A),,quad A=x:g(x)le y$$ Using graph, you are finding the region $A$.
– StubbornAtom
Jul 25 at 15:34
Oh, i see, thanks, but how could i find density function when i have this?
– cdummie
Jul 26 at 7:12
1
By differentiating the distribution function, you would get the density function, provided $Y$ is absolutely continuous.
– StubbornAtom
Jul 26 at 8:14
@StubbornAtom That works because differentiation is opposite of integration in a way, i think i got that, what about the second case b) , as far as i can see that is not continuous distribution, how am i supposed to act in that kind of situations, will it's distribution also be discrete?
– cdummie
Jul 27 at 18:53
1
No. $P(lfloor Xrfloor=k)=P(kle X<k+1)=cdots$
– StubbornAtom
Jul 28 at 15:13
2
2
This is the usual CDF method. If $Y=g(X)$, then distribution function of $Y$ is $$P(Yle y)=P(g(X)le y)=P(Xin A),,quad A=x:g(x)le y$$ Using graph, you are finding the region $A$.
– StubbornAtom
Jul 25 at 15:34
This is the usual CDF method. If $Y=g(X)$, then distribution function of $Y$ is $$P(Yle y)=P(g(X)le y)=P(Xin A),,quad A=x:g(x)le y$$ Using graph, you are finding the region $A$.
– StubbornAtom
Jul 25 at 15:34
Oh, i see, thanks, but how could i find density function when i have this?
– cdummie
Jul 26 at 7:12
Oh, i see, thanks, but how could i find density function when i have this?
– cdummie
Jul 26 at 7:12
1
1
By differentiating the distribution function, you would get the density function, provided $Y$ is absolutely continuous.
– StubbornAtom
Jul 26 at 8:14
By differentiating the distribution function, you would get the density function, provided $Y$ is absolutely continuous.
– StubbornAtom
Jul 26 at 8:14
@StubbornAtom That works because differentiation is opposite of integration in a way, i think i got that, what about the second case b) , as far as i can see that is not continuous distribution, how am i supposed to act in that kind of situations, will it's distribution also be discrete?
– cdummie
Jul 27 at 18:53
@StubbornAtom That works because differentiation is opposite of integration in a way, i think i got that, what about the second case b) , as far as i can see that is not continuous distribution, how am i supposed to act in that kind of situations, will it's distribution also be discrete?
– cdummie
Jul 27 at 18:53
1
1
No. $P(lfloor Xrfloor=k)=P(kle X<k+1)=cdots$
– StubbornAtom
Jul 28 at 15:13
No. $P(lfloor Xrfloor=k)=P(kle X<k+1)=cdots$
– StubbornAtom
Jul 28 at 15:13
 |Â
show 4 more comments
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2
This is the usual CDF method. If $Y=g(X)$, then distribution function of $Y$ is $$P(Yle y)=P(g(X)le y)=P(Xin A),,quad A=x:g(x)le y$$ Using graph, you are finding the region $A$.
– StubbornAtom
Jul 25 at 15:34
Oh, i see, thanks, but how could i find density function when i have this?
– cdummie
Jul 26 at 7:12
1
By differentiating the distribution function, you would get the density function, provided $Y$ is absolutely continuous.
– StubbornAtom
Jul 26 at 8:14
@StubbornAtom That works because differentiation is opposite of integration in a way, i think i got that, what about the second case b) , as far as i can see that is not continuous distribution, how am i supposed to act in that kind of situations, will it's distribution also be discrete?
– cdummie
Jul 27 at 18:53
1
No. $P(lfloor Xrfloor=k)=P(kle X<k+1)=cdots$
– StubbornAtom
Jul 28 at 15:13