# A very simple Expected Value question

Can someone explain to me why is $P(Y = 1) = P(X = -1)+P(X = 1)$?

Why is P(Y = 1) the sum of P(X = -1) and P(X= 1)? I don't see how $Y = X^2$ comes into play? I am very new to this stuff.

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We have $Y=1$ iff $X^2=1$ iff $X=1$ or $X=-1$.
The only way that $X^2$ can be $1$ is $X=1$ or $X=-1$. The probability that $X=1$ is $0.3$, the probability that $X=-1$ is $0.2$, so the probability that one or the other happens is $0.3+0.2$. You can draw a Venn diagram. Divide the "world" into $3$ non-intersecting pieces, label one of them $X=1$, another $X=-1$, and the third $X=0$. The probability that $X^2=1$ is the sum of the probabilities of the first two pieces.
So we are basically adding all possible values of $X$? What if $Y = X^3$? – Hawk Jul 28 '12 at 0:02
The only way that we can have $X^3=1$ is if $X=1$. So no addition needed. But for $X^4$, same issue as for $X^2$. An interesting one is $Y=X(X-1)$. Find probability $Y=0$. this can happen if $X=0$ or $X=1$, so it would be two other pieces you would add. – André Nicolas Jul 28 '12 at 0:07
Exactly the same, in this case as $E(X)$. This is because for the numbers $x=1$, $0$, $-1$, we have $x^3=x$. – André Nicolas Jul 28 '12 at 0:15