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I am trying to solve the following question from one of my classes.

I generally understand how to find the expected value and variance when given a random variable. However, when doing so, it is usually with an example where the values and probabilities of the random value are pretty easily defined (like rolling a die). In this example, I'm unsure of how to go about setting up and solving the problem because I'm not sure what values X or Y could take on and what the probabilities of those values would be. Could anyone show how to find expected values and variances in this problem?

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The question tells you exactly what values $X$ and $Y$ can take and what the probabilities are. $X$ takes the values $2$ and $-2$, each with probability $1/2$. $Y$ takes the values $4$ and $-1$, with probabilities $0.2$ and $0.8$.

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  • $\begingroup$ Ahh. This makes a lot more sense. This would mean the expected value and variance of X would be: E(X) =(2)(.5) + (-2)(.5) = 0 and E(X^2)= (2^2)(.5) + (-2^2)(.5) = 0 Would I then multiply these values based on how many shares were purchased? For example, 100 shares of A would be: 100 * E(X) = 0 and 100 * E(X^2) = 0 $\endgroup$ – user177772 Sep 22 '14 at 2:50
  • $\begingroup$ Your calculation of $E(X^2)$ is wrong: $(-2)^2 = 2^2$, not $-(2^2)$. Also notice that $(100 X)^2 = 100^2 X^2$, not $100 X^2$. $\endgroup$ – Robert Israel Sep 22 '14 at 4:47
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Someone asks you to play a game. He says

"I will throw a fair coin, if it comes up heads I will give you $2$ dollars if not, you will give me $2$ dollars. You can also choose to play with unfair coin which comes up heads $20 \% $ of the time. If it comes up heads I will give you $4$ dollars if not, you will give me $1$ dollar. "

In the fair coin case your expected profit: $$2*0.5+(-2)*0.5 = 0$$ In the unfair coin case: $$4*0.2+(-1)*0.8=0$$

It is better not to waste your time with him because your expected profit is $0$.

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