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Let

$$ Y =\sum_{k=1}^N a_kX_k $$

be the weighted sum of N independent random variables, $ X_k, k = 1, ... , N $ , each having mean $ \mu _{X_i} $ and variance $ \sigma ^2_{X_i} $. The weights $a_k$ are real-valued constants.

  • Derive an expression for the mean of $ Y, \mu _Y $
  • Derive an expression for the variance of $Y, \sigma ^2_Y $

Hint: Use the expectation operator...

Attempt:

$E[Y] = \mu _Y$ , or the expectation of $Y$ is the mean of $Y$. Since $Y$ represents the weighted sum of the N independent random variables, dividing $Y$ by the summation of the weights should give the mean, or:

$$ \mu _Y = E[Y] = Y / \sum_{k=1}^N a_k $$

where $Y$ is defined above.

Then we have $\sigma ^2_Y = E[Y^2] - \mu ^2_Y $ which can be derived from the above but gets very messy. Am I on the right track?

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    $\begingroup$ Hint: for $\mu_Y$ use the expectation operator on both sides of the equation for $Y$. You'll get $\mu_Y$ in terms of the $a_k$ and $\mu_X$. Don't forget about linearity of expectation... $\endgroup$ – Null Sep 8 '14 at 22:26
  • $\begingroup$ In your displayed equation for $\mu_Y$, the right side is a random variable (it is, in fact, just $Y$ scaled by a constant) while the left side is presumably a constant. So, clearly something is wrong. $\endgroup$ – Dilip Sarwate Sep 8 '14 at 22:35
  • $\begingroup$ @Null - This gives us $E[Y] = a_k * \sum_{k=1}^N X_k $ Since their means are given, is it then just the summation of the means $ \mu _X $ divided by N? $\endgroup$ – Ryde91s Sep 8 '14 at 22:35
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The expectation is a linear operator, so:

$$E[Y]=E\Biggl[\sum_{i=1}^Na_iX_i\Biggr]=\sum_{i=1}^Na_iE[X_i]=\sum_{i=1}^Na_i\mu_{X_i}$$

Then, you already know that $$V[Y]=E[Y^2]-E^2[Y]$$

So $$\begin{array}{rcl} E[Y^2]&=&E\Biggl[\Biggl(\sum_{i=1}^Na_iX_i\Biggr)^2\Biggr]\\ &=&E\Biggl[\sum_{i=1}^Na_i^2X_i^2+2\sum_{1\leq i<j\leq N}a_ia_jX_iX_j\Biggr]\\ &=&\sum_{i=1}^Na_i^2E\Bigl[X_i^2\Bigr]+2\sum_{1\leq i<j\leq N}a_ia_jE\Bigl[X_iX_j\Bigr]\\ \end{array}$$

You know that $E[X_i^2]=E^2[X_i]+V[X_i]=\mu_{X_i}^2+\sigma_{X_i}^2$, and $X_i$ are independent, so: $$\begin{array}{rcl} E[Y^2]&=&\sum_{i=1}^Na_i^2(\mu_{X_i}^2+\sigma_{X_i}^2)+2\sum_{1\leq i<j\leq N}a_ia_j\mu_{X_i}\mu_{X_j}\\ \end{array}$$

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It might be easier if you give yourself an example. Say each week you collect cans and bottles to recycle. Each can yields $0.05$ cents and each bottle yields $0.10$ cents. Let $X_1$ denote the amount of cans you find each week and $X_2$ the amount of bottles; on average you find $\mu_{X_1}$ cans and $\mu_{X_2}$ bottles. What's the average amount of money you make each week? Answer:

\begin{gather*} 0.05 \times \mu_{X_1} + 0.10 \times \mu_{X_2}. \end{gather*}

For the second part, note that \begin{gather*} E\left[\left(\sum_{i=1}^N X_i\right)^2\right] = E\left[\sum_{i=1}^N X_i^2 + 2\sum_{1 \leq i < j \leq N} X_i X_j\right]. \end{gather*} If $X_i$ and $X_j$ are independent, then $E[X_iX_j] = E[X_i]E[X_j]$.

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  • $\begingroup$ That makes sense with numerals, @snarski - but it gets lost in translation when it is abstracted for me. Would the expectation simply be $ \sum_{k=1}^N a_k * \mu _{X_k} $ ? $\endgroup$ – Ryde91s Sep 8 '14 at 22:54
  • $\begingroup$ Correct. Notice that what you did was $$E[Y] = E[0.05 X_1 + 0.10 X_2] = 0.05\mu_{X_1} + 0.10\mu_{X_2} = \sum_{i=1}^2\alpha_i\mu_{X_i},$$ where $\alpha_1 = 0.05$ and $\alpha_2 = 0.10$. $\endgroup$ – snar Sep 9 '14 at 0:53
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$$ \mu_y=\sum_{i=1}^k a_k E(X_k)=\mu \sum_{i=1}^k a_ix_i $$ Since $X_i$ are independent : $$ \sigma_y^2=\sum_{i=1}^k a_k^2 VAR(X_k)=\sigma^2 \sum_{i=1}^k a_i^2 x_i $$ EDIT Since $\mu$ and $\sigma$ have now the subscripts they should go under the sums. $$ \mu_y=\sum_{i=1}^k a_k E(X_k)=\sum_{i=1}^k a_i\mu_i $$ $$ \sigma_y^2=\sum_{i=1}^k a_k^2 VAR(X_k)=\sum_{i=1}^k a_i^2 \sigma^2_i $$

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  • $\begingroup$ The $\mu$ and $\sigma$ values were intended to have subscripts, not multipliers as was originally formatted which I believe makes this incorrect. $\endgroup$ – Ryde91s Sep 8 '14 at 22:52
  • $\begingroup$ Well, I added the note, but I cannot follow all changes you will decide to make. $\endgroup$ – Alexander Vigodner Sep 9 '14 at 1:05

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