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A fair die is tossed and let the random variable $X$ be the number that appears. Deduce that $$ \mathbb E(X^3)=\frac{1^3+2^3+3^3+4^3+5^3+6^3}6. $$ First of all, I would like to know the probability distribution of this random variable $X$.

Patrick Da Silva
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Junior II
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    We know what $X$ is, but what is $X_3$? – David Mitra Dec 22 '11 at 09:11
  • I edited X3 into $X_3$, maybe you intended $3X$? – Rudy the Reindeer Dec 22 '11 at 09:12
  • You really need to indicate what does your notation mean. First of all, we have no means to understand what does "13" stands for. We have even less understanding of what it means to "add" the things "xy" and "wz" by naming it "xy + wz". Perhaps it makes sense to you, and I'm sure it is not a very fancy thing (dice problems usually aren't very fancy), so just a little more detail would make your question more easy to answer. =)

    But comments apart, welcome to MSE =)

     – Patrick Da Silva Dec 22 '11 at 09:22
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    sorry, it is x cubed – Junior II Dec 22 '11 at 09:37
  • E(X3)=sum of X3f(x) for 1≤x≤6. here since it is a fair die should i put f(x)=1/6 for all x in [1,6]? – Junior II Dec 22 '11 at 09:51
  • If you want to cube things you can just add a "$^$" character in between the two things. I'll edit your question so that you can look at it. Where it says "edited n mins ago by", you can click on the link on "mins ago" to see what was edited. – Patrick Da Silva Dec 22 '11 at 10:06

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$X$ takes the values $1$, $2$, $3$ ,$4$, $5$, and $6$ (assuming a six-sided die). Since the die is fair, outcomes are equally likely; so $X$ takes the value $i$ with probability $1/6$ for each $i=1, 2,3,4,5,6$.

The probability distribution of $X$ is therefore $$ p_X(i)=\textstyle{1\over 6},\quad i=1, 2, \ldots, 6. $$

As a warm up to your problem, let's find $\Bbb E(X)$. The expected value of $X$ is

$$ \Bbb E(X)=\sum_{i=1}^6 \,i\, p_X(i)= 1\cdot {1\over6}+2\cdot {1\over6}+3\cdot {1\over6}+4\cdot {1\over6} +5\cdot {1\over6}+6 \cdot {1\over6}=3.5. $$

To find the expected value of a function of $X$, you could use the following fact:

Fact: For a discrete variable $X$ that takes the values $x_1, x_2,\ldots,x_n$, the expectation of a function $h(X)$ of $X$ is

$$\Bbb E\bigl(h(X)\bigr) =\sum_{i=1}^n h(x_i) p_X(x_i).$$

To find $\Bbb E(X^3)$, we apply the above fact with $h(x)=x^3$: $$ \Bbb E( X^3)=\sum_{i=1}^6\, i^3 p_X(i)= ( 1)^3\cdot {1\over6}+( 2)^3\cdot {1\over6}+( 3)^3\cdot {1\over6}+( 4)^3\cdot {1\over6} +( 5)^3\cdot {1\over6}+( 6)^3 \cdot {1\over6} . $$

David Mitra
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