## 数学代写|概率论代写Probability theory代考|STAT7203

2023年1月3日

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## 数学代写|概率论代写Probability theory代考|The distribution of a function of a random variable

Sometimes we know the distribution of a random variable, yet we would like to determine the distribution of a function of a random variable. If $X$ is a random variable, then, in order to determine the distribution of $Y=g(X)$, we first should recognize whether $Y$ is discrete or continuous by determining the possible values of random variable $Y$. If $Y=g(X)$ is a discrete random variable, then, in order to determine its distribution, we should obtain its probability function. On the other hand, if $Y=g(X)$ is a continuous random variable, then, in order to obtain its distribution, we should obtain its density function. To do so, firstly, we obtain the cumulative distribution function and then differentiate it. This results in the density function. For better comprehension of this issue, consider the following examples.

If $X$ is a continuous random variable with the density function of
$$f_X(x)=\frac{1}{10} ; 0<x<10,$$
then determine the probability function of random variable $Y=[X]$ (the sign [] denotes the integer part).

Solution. The random variable $Y=[X]$ can take on the values of 1 to 9 . Thus, $Y$ is a discrete random variable and in order to determine its distribution, we should find its probability function. Hence, we have: \begin{aligned} & P(Y=0)=P(0 \leq X<1)=\int_0^1 \frac{1}{10} \mathrm{dx}=\frac{1}{10} \ & P(Y=1)=P(1 \leq X<2)=\int_1^2 \frac{1}{10} \mathrm{dx}=\frac{1}{10} \ & \vdots \ & P(Y=9)=P(9 \leq X<10)=\int_9^{10} \frac{1}{10} \mathrm{dx}=\frac{1}{10} \end{aligned}
Therefore, as it is seen, the random variable $\mathrm{Y}$ adopts discrete values 0 to 9 with probability $\frac{1}{10}$.

## 数学代写|概率论代写Probability theory代考|Conditioning on continuous space

In Chapter 3 , we showed that, according to the law of total probability, if the should condition on possible parts of the space to calculate the probability of occurrence of that event. Now, if the probability of occurrence of an event depends on the result of a random variable, we should condition on its possible values. Consider the following example: The probability that a shooter’s shot hits the target is $\frac{1}{X^2}$ in which $X$ is the distance of the target from the shooter which is determined randomly and can take on $1,2,3,4$, and 5 with probability $\frac{1}{5}$. What is the probability that the shooter’s shot hits the target?

Solution. Since the probability that the shot hits the target depends on the distance from the target, we should condition on the distance from the target. Therefore, if event $\mathrm{E}$ denotes that the shot hits the target, we have:
$$P(E)=\sum_{x=1}^5 P(E \mid X=x) P(X=x)=\frac{1}{1} \times \frac{1}{5}+\frac{1}{4} \times \frac{1}{5}+\frac{1}{9} \times \frac{1}{5}+\frac{1}{16} \times \frac{1}{5}+\frac{1}{25} \times \frac{1}{5}=\frac{5269}{18000}$$
Now, if $X$ follows a continuous distribution taking on values in the interval $(1,5)$ with equal density $f_X(x)=\frac{1}{5}$, then the probability that the shot hits the target again depends on $X$ and we should condition on its possible values. However, herein, $X$ is a continuous random variable. In such cases, the general approach for conditioning is the same as before. We divide the possible values of the random variable $X$ into very small parts. Then, in contrast with discrete case that we denote the probability of each point by $P(X=x)$, for the continuous case, we denote the probability of each small interval of length $d x$ about point $x$ by $f_X(x) d x$. Finally, we can calculate the probability of occurrence of an event like $E$ for each small interval of the random variable $X$ and add them up as follows:
$$\left.P(E)=\int_1^5 P(E \mid X=x) f_X(x) d x=\int_1^5 \frac{1}{x^2} \times \frac{1}{5-1} d x=\frac{-1}{4 x}\right]_1^5=0.2$$

# 概率论代考

## 数学代写|概率论代写Probability theory代考|The distribution of a function of a random variable

$$f_X(x)=\frac{1}{10} ; 0<x<10,$$

$$P(Y=0)=P(0 \leq X<1)=\int_0^1 \frac{1}{10} \mathrm{~d} x=\frac{1}{10}$$

## 数学代写|概率论代写Probability theory代考|Conditioning on continuous space

$$P(E)=\sum_{x=1}^5 P(E \mid X=x) P(X=x)=\frac{1}{1} \times \frac{1}{5}$$

$$P(E)=\int_1^5 P(E \mid X=x) f_X(x) d x=\int_1^5 \frac{1}{x^2}$$

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