variance of product of independent random variables
Let $(X_n)_{ngeq1}$ be a sequence of i.i.d. random variables and $(Y_n)_{ngeq1}$ be another sequence of i.i.d random variables. Moreover, every $X_n$ is independent with every $Y_n$, for $n=1,2,....$.
$X_n$ is normally distributed with mean zero and variance of 2.
The density of $Y_n$ is $f(y)=2y$ for $0<y<1$ and zero otherwise.
What can I conclude about the variance of the product $Var(X_iY_i)=E[(X_i Y_i)^2]$?
expected-value
asked Dec 1 at 11:12
Matteo
23
add a comment |
Let $(X_n)_{ngeq1}$ be a sequence of i.i.d. random variables and $(Y_n)_{ngeq1}$ be another sequence of i.i.d random variables. Moreover, every $X_n$ is independent with every $Y_n$, for $n=1,2,....$.
$X_n$ is normally distributed with mean zero and variance of 2.
The density of $Y_n$ is $f(y)=2y$ for $0<y<1$ and zero otherwise.
What can I conclude about the variance of the product $Var(X_iY_i)=E[(X_i Y_i)^2]$?
expected-value
asked Dec 1 at 11:12
Matteo
23
1
You call $(X_i,Y_i)$ a sequence of iid random variables, but $(X_i,Y_i)$ is not a random variable. If $X_i,Y_i$ are random variables then $(X_i,Y_i)$ is a random vector. Do you mean to say that $X_i$ and $Y_i$ are independent for every $i$? And that secondly for every $i$ the vectors $(X_i,Y_i)$ are iid? If so, then what is the use of introduction of a sequence here? Finally, random variable $X$ and function $f$ seem to fall out of the sky. I suspect that $X_i$ has same distribution as $X$ and $f$ denotes PDF of distribution of $Y_i$, but that should at least be mentioned.
– drhab
Dec 1 at 11:26
See odelama.com/data-analysis/Commonly-Used-Math-Formulas
– MPW
Dec 1 at 11:27
I am sorry I was unclear, I come from econometrics and probably there are some conventions that are not strict for mathematicians. I modified the post, I hope is not clear. When I say "i.i.d" I mean every random variable is independent and identically distributed
– Matteo
Dec 1 at 13:46
add a comment |
Let $(X_n)_{ngeq1}$ be a sequence of i.i.d. random variables and $(Y_n)_{ngeq1}$ be another sequence of i.i.d random variables. Moreover, every $X_n$ is independent with every $Y_n$, for $n=1,2,....$.
$X_n$ is normally distributed with mean zero and variance of 2.
The density of $Y_n$ is $f(y)=2y$ for $0<y<1$ and zero otherwise.
What can I conclude about the variance of the product $Var(X_iY_i)=E[(X_i Y_i)^2]$?
expected-value
asked Dec 1 at 11:12
Matteo
23
Let $(X_n)_{ngeq1}$ be a sequence of i.i.d. random variables and $(Y_n)_{ngeq1}$ be another sequence of i.i.d random variables. Moreover, every $X_n$ is independent with every $Y_n$, for $n=1,2,....$.
$X_n$ is normally distributed with mean zero and variance of 2.
The density of $Y_n$ is $f(y)=2y$ for $0<y<1$ and zero otherwise.
What can I conclude about the variance of the product $Var(X_iY_i)=E[(X_i Y_i)^2]$?
expected-value
expected-value
asked Dec 1 at 11:12
Matteo
23
asked Dec 1 at 11:12
Matteo
23
edited Dec 1 at 13:48
asked Dec 1 at 11:12
Matteo
23
asked Dec 1 at 11:12
Matteo
23
asked Dec 1 at 11:12
Matteo
23
23
1
You call $(X_i,Y_i)$ a sequence of iid random variables, but $(X_i,Y_i)$ is not a random variable. If $X_i,Y_i$ are random variables then $(X_i,Y_i)$ is a random vector. Do you mean to say that $X_i$ and $Y_i$ are independent for every $i$? And that secondly for every $i$ the vectors $(X_i,Y_i)$ are iid? If so, then what is the use of introduction of a sequence here? Finally, random variable $X$ and function $f$ seem to fall out of the sky. I suspect that $X_i$ has same distribution as $X$ and $f$ denotes PDF of distribution of $Y_i$, but that should at least be mentioned.
– drhab
Dec 1 at 11:26
See odelama.com/data-analysis/Commonly-Used-Math-Formulas
– MPW
Dec 1 at 11:27
I am sorry I was unclear, I come from econometrics and probably there are some conventions that are not strict for mathematicians. I modified the post, I hope is not clear. When I say "i.i.d" I mean every random variable is independent and identically distributed
– Matteo
Dec 1 at 13:46
add a comment |
1
You call $(X_i,Y_i)$ a sequence of iid random variables, but $(X_i,Y_i)$ is not a random variable. If $X_i,Y_i$ are random variables then $(X_i,Y_i)$ is a random vector. Do you mean to say that $X_i$ and $Y_i$ are independent for every $i$? And that secondly for every $i$ the vectors $(X_i,Y_i)$ are iid? If so, then what is the use of introduction of a sequence here? Finally, random variable $X$ and function $f$ seem to fall out of the sky. I suspect that $X_i$ has same distribution as $X$ and $f$ denotes PDF of distribution of $Y_i$, but that should at least be mentioned.
– drhab
Dec 1 at 11:26
See odelama.com/data-analysis/Commonly-Used-Math-Formulas
– MPW
Dec 1 at 11:27
I am sorry I was unclear, I come from econometrics and probably there are some conventions that are not strict for mathematicians. I modified the post, I hope is not clear. When I say "i.i.d" I mean every random variable is independent and identically distributed
– Matteo
Dec 1 at 13:46
1
1
You call $(X_i,Y_i)$ a sequence of iid random variables, but $(X_i,Y_i)$ is not a random variable. If $X_i,Y_i$ are random variables then $(X_i,Y_i)$ is a random vector. Do you mean to say that $X_i$ and $Y_i$ are independent for every $i$? And that secondly for every $i$ the vectors $(X_i,Y_i)$ are iid? If so, then what is the use of introduction of a sequence here? Finally, random variable $X$ and function $f$ seem to fall out of the sky. I suspect that $X_i$ has same distribution as $X$ and $f$ denotes PDF of distribution of $Y_i$, but that should at least be mentioned.
– drhab
Dec 1 at 11:26
You call $(X_i,Y_i)$ a sequence of iid random variables, but $(X_i,Y_i)$ is not a random variable. If $X_i,Y_i$ are random variables then $(X_i,Y_i)$ is a random vector. Do you mean to say that $X_i$ and $Y_i$ are independent for every $i$? And that secondly for every $i$ the vectors $(X_i,Y_i)$ are iid? If so, then what is the use of introduction of a sequence here? Finally, random variable $X$ and function $f$ seem to fall out of the sky. I suspect that $X_i$ has same distribution as $X$ and $f$ denotes PDF of distribution of $Y_i$, but that should at least be mentioned.
– drhab
Dec 1 at 11:26
See odelama.com/data-analysis/Commonly-Used-Math-Formulas
– MPW
Dec 1 at 11:27
See odelama.com/data-analysis/Commonly-Used-Math-Formulas
– MPW
Dec 1 at 11:27
I am sorry I was unclear, I come from econometrics and probably there are some conventions that are not strict for mathematicians. I modified the post, I hope is not clear. When I say "i.i.d" I mean every random variable is independent and identically distributed
– Matteo
Dec 1 at 13:46
I am sorry I was unclear, I come from econometrics and probably there are some conventions that are not strict for mathematicians. I modified the post, I hope is not clear. When I say "i.i.d" I mean every random variable is independent and identically distributed
– Matteo
Dec 1 at 13:46
add a comment |
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1
You call $(X_i,Y_i)$ a sequence of iid random variables, but $(X_i,Y_i)$ is not a random variable. If $X_i,Y_i$ are random variables then $(X_i,Y_i)$ is a random vector. Do you mean to say that $X_i$ and $Y_i$ are independent for every $i$? And that secondly for every $i$ the vectors $(X_i,Y_i)$ are iid? If so, then what is the use of introduction of a sequence here? Finally, random variable $X$ and function $f$ seem to fall out of the sky. I suspect that $X_i$ has same distribution as $X$ and $f$ denotes PDF of distribution of $Y_i$, but that should at least be mentioned.
– drhab
Dec 1 at 11:26
See odelama.com/data-analysis/Commonly-Used-Math-Formulas
– MPW
Dec 1 at 11:27
I am sorry I was unclear, I come from econometrics and probably there are some conventions that are not strict for mathematicians. I modified the post, I hope is not clear. When I say "i.i.d" I mean every random variable is independent and identically distributed
– Matteo
Dec 1 at 13:46