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Calculate the expected value

mathmari

Well-known member
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Apr 14, 2013
4,036
Hey!! :eek:

The random variables $X_1, X_2, \ldots , X_{10}$ are independent and have the same distribution function and each of them gets exactly the values $\pm 2$ and with equal probability.

We define the random variable $S=X_1+X_2+\ldots +X_{10}$.

I want to calculate $\mathbb{E}(S^2)$.

Could you give me a hint how we could calculate that? I don't really have an idea. (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
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Hey mathmari!! (Smile)

What is the expectation of $X_1$?
Of $X_1+X_2$?
Of $X_1^2$? (Wondering)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
What is the expectation of $X_1$?
Of $X_1+X_2$?
Of $X_1^2$? (Wondering)
Do we have the following?

$\mathbb{E}(X_i)=x_i\cdot p=(\pm 2)\cdot \frac{1}{10}=\pm\frac{1}{5}$
$\mathbb{E}(X_i+X_j)=\mathbb{E}(X_i)+\mathbb{E}(X_j)$
$\mathbb{E}(X_i^2)=x_i^2\cdot p=(\pm 2)^2\cdot \frac{1}{10}=4\cdot \frac{1}{10}=\frac{2}{5}$

(Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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Don't we have:
$$EY=\sum_j y_jp_j$$
(Wondering)
 

mathmari

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Apr 14, 2013
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Klaas van Aarsen

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Mar 5, 2012
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So do we have $E(X_i)=-2\cdot \frac{1}{2}+2\cdot \frac{1}{2}=0$ ? Or do you mean something else?
Yes, that's what I meant.
 

mathmari

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Apr 14, 2013
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Klaas van Aarsen

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Mar 5, 2012
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Great!! Do we get then $$E(S^2)=\sum E(X_i^2)=\sum x_i^2\cdot p=\sum 4\cdot \frac{1}{2}=10\cdot 2=20$$?
Isn't $S^2\ne \sum X_i^2$? (Worried)
 

mathmari

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Apr 14, 2013
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Klaas van Aarsen

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Mar 5, 2012
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Ah ok.. but what can we do in this case?
I see the following possible approaches:

  1. Apply the definition of expectation directly.
    $$E(S^2) = \sum_j s_j^2 q_j$$
    where $s_j$ is each of the possible $n^2$ outcomes and $q_j=\left(\frac 12\right)^2$ are the corresponding probabilities.

  2. Use the calculation rules that apply to expectations:
    $$E(S^2) = E\Big((X_1 + .. + X_n)^2\Big) = E\Big(X_1^2 + .. X_n^2 + \sum_{i\ne j} 2X_iX_j\Big) = E(X_1^2) + ... + E(X_n^2) + 2 \sum_{i\ne j} E(X_iX_j)$$
    What is $E(X_iX_j)$ with $i\ne j$?

  3. Use that generally $\sigma^2(Y) = E\Big((Y-EY)^2\Big) = E(Y^2) - (EY)^2$ and substitute $Y=S=X_1+...+X_n$.
(Thinking)
 

Dhamnekar Winod

Active member
Nov 17, 2018
103
I see the following possible approaches:

  1. Apply the definition of expectation directly.
    $$E(S^2) = \sum_j s_j^2 q_j$$
    where $s_j$ is each of the possible $n^2$ outcomes and $q_j=\left(\frac 12\right)^2$ are the corresponding probabilities.

  2. Use the calculation rules that apply to expectations:
    $$E(S^2) = E\Big((X_1 + .. + X_n)^2\Big) = E\Big(X_1^2 + .. X_n^2 + \sum_{i\ne j} 2X_iX_j\Big) = E(X_1^2) + ... + E(X_n^2) + 2 \sum_{i\ne j} E(X_iX_j)$$
    What is $E(X_iX_j)$ with $i\ne j$?

  3. Use that generally $\sigma^2(Y) = E\Big((Y-EY)^2\Big) = E(Y^2) - (EY)^2$ and substitute $Y=S=X_1+...+X_n$.
(Thinking)
Hello,
What is $E(X_iX_j)$ with $i\ne j$? would you explain?
 

Klaas van Aarsen

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Mar 5, 2012
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Hello,
What is $E(X_iX_j)$ with $i\ne j$? would you explain?
An expectation is the sum of the possible outcomes times their probability.
In this case the possible outcomes are $\pm2 \cdot \pm 2$ and since they are independent each has probability $\frac 12 \cdot \frac 12 = \frac 14$.
So:
$$E(X_iX_j) = (-2\cdot -2)\cdot \frac 14 + (-2 \cdot 2)\cdot \frac 14 + (2 \cdot -2) \cdot \frac 14+ (2\cdot 2)\cdot \frac 14 = 0$$