Linked Questions
15 questions linked to/from Variance of product of multiple independent random variables
6
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1
answer
68k
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Var(XY), if X and Y are independent random variables [duplicate]
if X and Y are independent Random variable then what is the variance of XY?
0
votes
0
answers
3k
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How to calculate variance or standard deviation for product of two normal distributions? [duplicate]
For example if I have two multiplied distributions a * b:
...
0
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0
answers
2k
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Define $Var(XY)$ in terms of $E(X)$, $E(Y)$, $Var(X)$, $Var(Y)$ for Independent Random Variables $X$ and $Y$ [duplicate]
I currently know: $$Var(XY)=E(X^2Y^2)−[E(XY)]^2$$
$$=E(X^2)E(Y^2)−[E(X)]^2[E(Y)]^2$$
but I am lost where to go from there. I can vaguely see the the formula $Var(X)=E(X^2)-E(X)^2$ hidden somewhere, ...
40
votes
2
answers
62k
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Variance of product of dependent variables
What is the formula for variance of product of dependent variables?
In the case of independent variables the formula is simple:
$$ {\rm var}(XY) = E(X^{2}Y^{2}) - E(XY)^{2} = {\rm var}(X){\rm var}(...
14
votes
3
answers
6k
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Variance of product of k correlated random variables
What is the variance of the product of $k$ correlated random variables?
11
votes
1
answer
6k
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Variance of powers of a random variable
Is it possible to derive a formula for variance of powers of a random variable in terms of expected value and variance of X?
$$\operatorname{var}(X^n)= \,?$$
and
$$E(X^n)=\,?$$
4
votes
1
answer
528
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Standard deviation/variance for the sum, product and quotient of two Poisson distributions
What would be the standard deviation for $A+B$, $AB$ and $\frac{A}{B}$ for $A$ and $B$ Poisson distributed?
1
vote
2
answers
511
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If you know a normal distribution's population variance, does a sample variance tell you nothing about the sample's mean's confidence interval?
If I understand correctly (which I might not), if I know a normal distribution's population variance but not its population mean, and take just one sample consisting of three measurements, then no ...
5
votes
1
answer
147
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Point estimator for product of independent RVs
Let $X$ and $Y$ be two independent random variables. Given an (iid) random sample of size $n$ of $X$ and a random sample of size $n$ of $Y$, what is a good way to estimate the mean of their product, $...
4
votes
1
answer
285
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Why is estimating the standard error of an estimate that is itself the product of several estimates so difficult?
Singer and Willett (2003) write the following about estimating the standard errors of estimated survival probabilities within the context of discrete time event history models (e.g. logit hazard ...
0
votes
2
answers
170
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Variance of the difference of products of iid sequences
So suppose you have two sequences $\{Y_t\}$ and $\{Z_t\}$ and they are both iid and independent from each other. Now suppose I have a time series $\{X_t\}$ such that...
$$\{X_t\} = Y_t(1-Y_{t-1})Z_t$$...
3
votes
2
answers
162
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What is $V(X^t)$ for any $t$ when only $E(X)$ and $Var(X)$ are known and $X$ is assumed normal?
Summary
I'm trying to calculate $Var(X^t)$ where $t$ is the number of periods using only the following known parameters:
$E(X)$ and $Var(X)$. $X$ is a random variable and is the return factor $(1 + ...
2
votes
0
answers
189
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Is the product of two Wrapped Normal Variables a Wrapped Normal Variable? [closed]
I know that the sum of two Wrapped Normal Variables is a Wrapped Normal Variable. In particular, if $\theta_1 \sim WN(\mu_1, \rho_1)$ and $\theta_2 \sim WN(\mu_2, \rho_2)$, then $\theta_1 + \theta_2 \...
5
votes
0
answers
112
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How can I calculate the probability that the product of two independent random variables does not exceed $L$?
I have one variable, $X$, which is provided hourly for a period of one month (720 total values in the series). I have another variable, $Y$, which is provided quarterly (for which I am provided the ...
0
votes
0
answers
87
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Finding the best predictor Brownian motion
I want to find the best predictor of $(B_3-B_2)(B_4-B_{\pi})$ given an observation of $B_1$
Where $B_t$ is brownian motion for time $t \geq 0$.
I am not sure how to approach this.
I know it will be ...