The above is from my prob book. Could you please explain what is the reason to switching to z from y in the above case. It might be a basic question, but would appreciate any help and detailed explanation.
Thank you.
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1$\begingroup$ Please do not use pictures for critical portions of your post. Pictures cannot be searched and are inaccessible to those using screen readers. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 13 '17 at 22:21
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Basically, the index used by series notation is just a token. You can interchangeably use any symbol that is otherwise free in the term (basically: that is not already used).
$$\sum_z g(x,z) ~=~ \sum_y g(x,y) ~\neq~ \sum_x g(x,x)$$
So in this case, the author choose to use $z$ when introducing a bound variable for a series, because the symbol is otherwise free in the expression. In particularly, when $y$ needs to be a boundary condition, it can not also be a bound variable. They could have chosen $\beta$, or most anything, really.
$$F_{\lower{0.5ex}{Y\mid X=x}}(\color{blue}y) = \sum_{\beta\leq \color{blue}y}p_{\lower{0.5ex}{Y\mid X=x}}(\beta)$$
answered Mar 14 '17 at 0:03
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This is just to distinguish between the argument of the function and the index you are summing over. That is a function (with finite domain) $$f(y) = \sum_{z\leq y} g(z) $$ means that you will sum over all values that are less than or equal to $y$. That is $$ f(y) = \underbrace{g(value1)}_{z=value1} + \underbrace{g(value2)}_{z=value2} + \dots + \underbrace{g(lastValue)}_{z=lastValue \leq y}$$ in which $$ value1 < value2 < \dots < lastValue \leq y$$.
You also do the same for integrals. For example $$f(y) = \int_{-\infty}^y g(z) dz$$ Here we use $z$ which is a dummy variable (that we integrate/sum over) to be able to use $y$ as the argument of the function.
