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"If there is uncertainty about some monetary outcome and you are concerned about return and risk, then all you need to know are the mean and standard deviation of the possible outcomes. The entire distribution provides no extra useful information". Do you agree or disagree? Justify your answer and provide an example to back up your argument.

I disagree. If $\text{ return } \sim \mathcal{U}[0,4]$we know the mean is $2$ and standard deviation $\sqrt{0.3}$ of the return, it is different than having a normal distribution with such parameters as the uniform would give us minimum $0$ return, whereas if the return is normally distributed, it could give us negative return. So knowing the distribution does make a difference.

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  • $\begingroup$ The standard deviation is actually $\sqrt{\frac13}$, but apart from that, you have given a reasonable example. You might add the normal probability of being negative to quantify your point $\endgroup$
    – Henry
    Jul 29, 2021 at 9:36

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I give you another example to enforce your good conclusion.

Suppose you have this two random return rv's:

$X=\{0;10\}$ and $Y=\{-1;1\}$ and any occurrence's probability equal $0.5$.

If you calculate mean and variance of the two rv's you get

X with Mean 5 and Variance 25

Y with Mean 0 and Variance 1

Now, using the mean variance method, you cannot decide which project is the best one; infact X has higher mean but higher risk w.r.t Y

BUT, if you look at the two invesments, X gives you only the probabilty to win, while Y gives you 50% probabilty to loose. Moreover, with the same 50% probability, with project X you win 10 while with Y you can only win 1.

Thus it is self evident that $X$ is preferred to $Y$ but you can take this decision if you have the distribution, not only with the two indicators, $\mu$ and $\sigma$

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