fundamental statistics theory notes (1)

Suppose y is a continuous random variable, which has a PDF (Probability distribution function) $f(y)$ and a CDF (Cumulative distribution function) $F(y)$ shown as below

$$P(a<=y<=b)=int_a^bf(y)dy$$

Generally

$$int_{-infty}^{+infty}{f(y)dy}=1$$

For CDF, $F(y)$ is equal to $P(Y<=y) = int_{-infty}^{y}f(t)dt$. If we differtiate the $F(y)$, we will get $f(y)$, the PDF, like $f(y) = frac{d}{dy}F(y)$.

Example:
Suppose $zsim N(0,1)$, $f(z) = frac{1}{sqrt{2pi}}e^{frac{-z^2}{2}}$, $z epsilon R$.
Define $Y = |z|$, find PDF/CDF of Y.
First Step: Identify the range
$Ygeq0$
For any y greater than 0, $P(Yleq y) = P(|z|leq y) = P(-yleq zleq y) = int_{-y}^{y}frac{1}{sqrt{2pi}}e^{frac{-z^2}{2}}dz$
Define $Phi(z) = $ CDF of $N(0,1) = int_{-infty}^zfrac{1}{sqrt{2pi}}e^{frac{-z^2}{2}}dz Rightarrow Phi(y) - Phi(-y)$, which is the CDF of y for any y $geq0$.

Second Step: Get derivative of CDF
PDF of Y is $ f(y) = frac{d}{dy} left( Phi(y) - Phi(-y) right) $
= $ Phi’(y) + Phi’(-y) $
= $frac{1}{sqrt{2pi}} e^{frac{(-y)^2}{2}}$ + $frac{1}{sqrt{2pi}} e^{-{frac{(-y)^2}{2}}} $
= $frac{2}{sqrt{2pi}} e^{-frac{y^2}{2}} $ for any y $geq 0$

For Y contains with PDF $f(y)$
$E(Y) = int_{-infty}^{+infty}yf(y)dy$ which is “expected value” or “mean”.
$E(g(y)) = int_{-infty}^{+infty}g(y)f(y)dy$
Varince: $Var(Y) = E(Y^2) - E(Y]^2$

• Expectations are linear: $E(a + bx + cy) = a + bE(x) + cE(y)$
• When x, y are independent: $E[g(x)h(y)] = E[g(x)]E[h(y)]$
• $M_y(t) = E[e^{ty}]$ is the MGF of Y. The MGF uniquely determines the destribution, i.e. if $M_x(t) = M_y(t)$ then $x$~$y$.

Example:
$xsim Expo(beta)$, which has MGF $M_x(t) = frac{1}{1-beta t}$. Whta is the distribution of $Y=2x$?

The MGF of Y is $M_y(t) = E[e^{ty}] = E[e^{tcdot2x}] = E[e^{(2t)x}] = M_x(2t) = frac{1}{1-2beta t}$, which is the MGF of $Expo(2beta)$.

Since MGF determines the distribution, $Ysim Expo[2beta]$.