I am trying to calculate the first two moments of the random walk below:
$y_t$=$y_0$+$\sum_{j=1} ^t u_j$
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Mean: E[$y_t$]=$y_0$
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Variance: E[$y_t ^2$]=t$\sigma^2$
I understand how to obtain the 1st moment because we assume the expectation of the disturbance terms is equal to 0 and the expectation of a constant is the constant.
I do not understand how to obtain the variance. I tried the following formula:
var($y_t$)=E[$y_t ^2$]-$E[y_t]^2$
I incorrectly simplified this in the following steps:
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$var(y_t)=E[(y_0+\sum u_j)(y_0+\sum u_j)]+y_0 ^2$
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$var(y_t)=E[y_0 ^2+2\sum u_j+\sum u_j ^2]+y_0 ^2$
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$var(y_t)=E[y_0]+2E[\sum u_j]+E[(\sum u_j)^2]$
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$var(y_t)=y_0$
I am hoping someone can explain that form of the variance is obtained.
This information comes from the ninth slide of the following presentation:
http://www.ncer.edu.au/events/documents/Lecture1_Intro.pdf
Best Answer
For the random walk $y_t = y_0 + \sum u_j$ where disturbance terms $u_j$s are modeled as white noise, mean can be calculated as: $E[y_t] = E[y_0 + \sum u_j] = E[y_0]+ E[\sum u_j] = E[y_0] + \sum E[u_j] = y_0 + \sum 0 = y_0$.
If var($y_t$) = E[$y_t^2$] - $E[y_t]^2$ formula is used to calculate variance of $y_t$ some simplifications can be applied as in the following steps:
Since $u_j$~$WN(0,\sigma ^2)$, $u_j$ terms are uncorrelated (all covariances of them except variances will be zero). Therefore, expectations in step 4. can be replaced with following values:
If these substitutions are applied, rest of the simplification steps can be written as follows: