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# 统计代写|统计推断代考Statistical Inference代写|STA2302 Discrete conditional distributions

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## 统计代写|统计推断代考Statistical Inference代写|Discrete conditional distributions

Consider discrete random variables $X$ and $Y$. Suppose we know that $X$ takes some particular value, $x$. This knowledge will affect our assessment of the distribution of probability associated with $Y$. We return to an example from Chapter 4 to illustrate.
Example 5.1.1 (Card drawing example again)
Recall in Example 4.2.1 we draw two cards at random without replacement from a deck of 52 and define $X$ to be the number of kings, and $Y$ to be the number of aces. Suppose we are told that exactly one king has been drawn, that is, $X=1$. This will affect our view of the distribution of the number of aces. The most obvious immediate consequence is that we now know that there cannot be two aces. We can work out the other probabilities using our knowledge of conditional probability and the results from Table 4.1.
\begin{aligned} \mathrm{P}(0 \text { aces } \mid 1 \text { king }) & =\mathrm{P}(Y=0 \mid X=1)=\frac{\mathrm{P}(X=1, Y=0)}{\mathrm{P}(X=1)} \ & =\frac{f_{X, Y}(1,0)}{f_X(1)}=\frac{0.1327}{0.1448}=0.916, \ \mathrm{P}(1 \text { aces } \mid 1 \text { king }) & =\mathrm{P}(Y=1 \mid X=1)=\frac{\mathrm{P}(X=1, Y=1)}{\mathrm{P}(X=1)} \ & =\frac{f_{X, Y}(1,1)}{f_X(1)}=\frac{0.0121}{0.1448}=0.084, \ \mathrm{P}(2 \text { aces } \mid 1 \text { king }) & =\mathrm{P}(Y=2 \mid X=1)=\frac{\mathrm{P}(X=1, Y=2)}{\mathrm{P}(X=1)} \ & =\frac{f_{X, Y}(1,2)}{f_X(1)}=\frac{0.0000}{0.1448}=0.000 . \end{aligned}
Note that $\mathrm{P}(0$ aces $\mid 1$ king $)+\mathrm{P}(1$ aces $\mid 1$ king $)+\mathrm{P}(2$ aces $\mid 1$ king $)=1$.

## 统计代写|统计推断代考Statistical Inference代写|Continuous conditional distributions

For continuous random variables, we define the conditional density to have the same form as the conditional mass function.
Definition 5.2.1 (Conditional density)
Suppose that $X$ and $Y$ are jointly continuous random variables with joint density $f_{X, Y}$ and that the marginal density of $X$ is $f_X$. The conditional density of $Y$ given $X=x$ is defined as
$$f_{Y \mid X}(y \mid x)= \begin{cases}\frac{f_{X, Y}(x, y)}{f_X(x)} & \text { for } f_X(x)>0 \ 0 & \text { otherwise. }\end{cases}$$

We cannot interpret conditional density as a conditional probability; recall that, if $X$ is continuous, then $\mathrm{P}(X=x)=0$ for any real number $x$. However, we can readily show that, viewed as a function of $y$, the function $f_{Y \mid X}(y \mid x)$ is a valid density function. This is part of Exercise 5.2 .

We define $Y \mid X=x$ as the random variable with density function $f_{Y \mid X}(y \mid x)$. As we might expect, the cumulative distribution function and expected value of $Y \mid X=x$ are found by integration,
\begin{aligned} F_{Y \mid X}(y \mid x) & =\int_{-\infty}^y f_{Y \mid X}(u \mid x) d u, \text { and } \ \mathbb{E}(Y \mid X=x) & =\int_{-\infty}^{\infty} y f_{Y \mid X}(y \mid x) d y . \end{aligned}

# 统计推断代写

## 统计代写|统计推断代考Statistical Inference代写|Discrete conditional distributions

$$\mathrm{P}(0 \text { aces } \mid 1 \text { king })=\mathrm{P}(Y=0 \mid X=1)=\frac{\mathrm{P}(X=1, Y=0)}{\mathrm{P}(X=1)} \quad=\frac{f_{X, Y}(1,0)}{f_X(1)}=\frac{0.1327}{0.1448}=0.916, \mathrm{P}(1 \text { aces } \mid 1 \text { king })$$

## 统计代写|统计推断代考Statistical Inference代写|Continuous conditional distributions

$$f_{Y \mid X}(y \mid x)=\left{\frac{f_{X Y}(x, y)}{f X(x)} \quad \text { for } f_X(x)>00 \quad\right. \text { otherwise. }$$

$$F_{Y \mid X}(y \mid x)=\int_{-\infty}^y f_{Y \mid X}(u \mid x) d u, \text { and } \mathbb{E}(Y \mid X=x) \quad=\int_{-\infty}^{\infty} y f_{Y \mid X}(y \mid x) d y$$

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## MATLAB代写

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