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## Useful formulas
> $$\mathrm{E}[XY]=\sum_{x_k=1}^{x_n} \sum_{y_k=1}^{y_n} x_k*y_k*\mathrm{P}(X=x_k,Y=y_k)$$
> $$\mathrm{E}[f(X) \mid Y=y_0]=\sum_{x_k=1}^{x_n} f(X=x_k)*\mathrm{P}(X=x_k \mid Y=y_0)$$

> $A^c \cup B^c = 1-A \cap B$



> let $I_k$ be an indicator, $\mathrm {E[I_k]=\sum_{k=0}^{1} k * P(I_k=k)=1*P(I_k=1)=P(I_k=1)}$
$\mathrm {V[I_k]=\sum_{k=0}^{1} k * P(I_k=k)=1*P(I_k=1)=P(I_k=1)}$

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## Statistics

### Find Ultimate test when $λ(x)$ is monnotone and rising and $\lim_{x \to \infty}\lambda(x)=\infty$
> -  Use $C^{c}$ instead of $C$ to find $\alpha$.
> 1. Use the following equation: $\alpha=P_(h_0)(C)=1-P_(h_0)(C^{c})=1-P(x_{1} + x_{2} + \cdots + x_{n}= min_0,..,max_0) = 1 -\sum_{k=min_0}^{max_0}P(x_{1} + x_{2} + \cdots + x_{n}=k)$
> - Explaination
    - Since $C$ contians **infinite** amount of numbers it will be easier to calculate what goes into $C^{c}$ which is exactly what this formula is used for
> - Usage
 >    - $min_0$ is the minimal value that can be applied to the sum of $x_{1} + x_{2} + \cdots + x_{n}$
 >    - $max_0$ is the **current** value that can be applied to the sum of $x_{1} + x_{2} + \cdots + x_{n}$ (for example we might be checking if $3$ **is the last number** that can be inserted into $C^{c}$ in which case we plug $max_0=3$)
 >     - we begin by setting $max_0$ to $min_0 +1$ and use the formula to find $\alpha$.
 >        - if $\alpha$ reached the limit stop
 >        - else increment $max_0$ by 1 and continue using and repeat the step above
 >    - once $\alpha$ reaches the limit we can define:
 >        - $C=\{max_0 \le x_{1} + x_{2} + \cdots + x_{n}\}$
 >        - $C^{c}=\{min_0 \le x_{1} + x_{2} + \cdots + x_{n} \le max_0\}$


### Finding maximal power test 
> for $H_0, H_1$ when $X_1=x_1,...,X_n=x_n andX=\sum{k=1}^{n}Xk$ are finite.
> 1. Define $P(X_1=x_1,...,X_n=x_n)$ for $H_0$ and $H_1$ seperately
> 2. If $X$ has a "few" possible values, define a chart as follows

| X  | $h_0: P(X_1=x_1,...,X_n=x_n)$ | h_1: P(X_1=x_1,...,X_n=x_n) | $\lambda(x)$ |
| ------------- | ------------- | ------------- | ------------- |
| $k$  | sum of all $h_0: P(X_1=x_1,...,X_n=x_n)$ when $x_1,...,x_n = k$ | sum of all $h_1: P(X_1=x_1,...,X_n=x_n)$ when $x_1,...,x_n = k$  | ratio of the previous 2 cells $\frac{h_1}{h_0})$
| $N$  | sum of all $h_0: P(X_1=x_1,...,X_n=x_n)$ when $x_1,...,x_n = N$ | sum of all $h_1: P(X_1=x_1,...,X_n=x_n)$ when $x_1,...,x_n = N$  | ratio of the previous 2 cells $\frac{h_1}{h_0})$



### מציאת אומדן עבור אומד
> $T(X_{1}, X_{2},...,X_{n})$ for instance $T = \frac{X^{c}-5}{10}$

> $X_1$=$x_{1}$, $X_2=x_{2}$, ... $X_n=x_{n} = M$
> :ועבור מדגם

> plug $M$ to T
> $T(M)=\frac{x_{1}+x_{2}+ \cdots x_{n}-5}{10}$
>Finding א.ח.ה for $\theta$ when we have sample $M= x_{1}, x_{2},...,x_{n}$:
>1. First we define $X_T = \frac{\sum_{k=1}^{n}x_k}{n}$ -- this will provide useful information as to what we need to change in order for $X_T$ to equal $\theta$, this changes on $X_T$ are $T$.
 >    - Useful tip: Avoid using the index ($k$) in T and opt for constant values. This will provide useful if we need to determine $MSE(\theta,T)$
> 2. Define $T$ based on the above "intuition" and then prove that these changes are sufficient by calculating $E[T]$


## Combinatorics:
> - $\binom{n}{k}$ = choose k items from population of n with no account towards the order of selection (no external order)
> - $\binom{n}{k}*k!$ = choose k items from population of n with account to the order of selection (external order)
> - The amount of groups with the size of $k$ items from the population of $n$ is $n-k+1$
> - The count of distributions of $n$ **different** items into a circle where the cells are **unordered** is $(n-1)!$
> - The count of distributions of $n$ **identical** items into $r$ numbered cells, where each cell must contain at least 1 item is: $\binom{n-1}{r-1}=\binom{n-1}{n-r}$
> - The count of distributions of $n$ **identical** items into $r$ numbered cells is: $\binom{n-1+r}{n}$
### Stategies
> - We generally prefer to not include external order in our selection to avoid duplicates
> - Use specific use-cases and draw them (Venn diagram/other illustrations) to make sure we count all necessary items and only once
> - Keywords like "at-least", "at-most" indicate there could be a couple of overlapping events. If the above condition excludes only 1 case, we should try counting using a complement
 >    - If there are more than 1 distinguishable cases we should use the **principle of inclusion-exclusion** to avoid counting duplicates.
> - With complex problems we should try to reduce the problem at hand into the lowest level problem that can be easily solved and then work our way up. For example: **indicators** often rely on this exact strategy
 >    - This strategy also includes separating the problem into separate stranger events where we can count each of them on it's own.
## 2D Discrete Probability Function of X,Y
> - if X,Y both have a small finite amount of cases we should draw a chart to describe $P(X=x, Y=y)$
 >    - when calculating the unified function, it could prove useful calculating $P(X=x)$ and $P(Y=y)$ to understand what results we expect the unified function $P(X=x,Y=y)$ to reach
> - else we should find the common function using: combinatorics/special distribution etc
> - sometimes X or Y's definition is dependent on the other and it could be that the dependent variable has a special distribution, in such cases, (assuming Y depends on X), given $X = r$ we can say $T=Y|X=r \sim SpecialDistribution(X,...)$

## Special distributions
>- The relationship between Exponential and Poesonic distributions is that $X \sim {EXP}(\frac{1}{\lambda})$ and $Y \sim {Po}(\lambda t)$ is t that $Y$ counts the amount of an event occurances within [0,T] time (for example 1 hour) and $X$ measures the waiting time to the next event (still time).
 >    - Count of events within T time$=N \sim {Po}(\lambda t)$ $\rightarrow$ waiting time until the next event$=X \sim {EXP}(\lambda)$
