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I saw in a statistics book a problem.

Let $X$ be a distribution that gets $1$ for probability $0.4$ and $2$ for probability $0.6$. Compute the mean and variances of $Y=3X-2$ and $Y=3X^2-2$. I found formulas and solved the problem. But how one defines the distributions $3X-2$ and $3X^2-2$? Because I found that $3\cdot 0.4-2+4\cdot 0.6-2=1.2-2+2.4-2=-0.6\ne 1$.

I have read mathematics so I'm not afraid to see a formal definition of the distribution or how to transform them.

eric_kernfeld
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    Welcome to the site! In the future, please add the self-study tag for this type of question. (I did it for this one.) – eric_kernfeld Dec 08 '17 at 16:32
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    Your description of $X$ is that of a *random variable* rather than a distribution. Our posts about random variables may help you understand better what expressions like $3X-2$ and $3X^2-2$ mean. Look at https://stats.stackexchange.com/questions/50 and https://stats.stackexchange.com/questions/95993. The question at https://stats.stackexchange.com/questions/140060/cdf-of-the-function-of-a-random-variable is a generalized version of yours, couched in terms of distribution functions (CDFs). – whuber Dec 08 '17 at 17:23
  • The random variable X is a discrete random variable. X has mean 0.4 + 1.2 =1.6 and its variance is 0.4 (1 -1.6)$^2$+0.6 (2-1.6)$^2$=0.144+0.96=0.24. In the first case Y is linear function of X so E(Y)=3 E(X)-2. Var(Y)=9 Var(X). – Michael R. Chernick Dec 08 '17 at 17:46
  • The second case is more complicated. But another way to get the moments of Y would be first determine the distribution of Y and directly calculate its mean and variance in the same way I did for X. In your question it is not clear what you are computing when you say -0.6 is not equal to 1. – Michael R. Chernick Dec 08 '17 at 17:53

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