台大應數所(碩士班) 110年度一般考試機率統計解答

1. Let $T$ be a failure time with support $\{t_1, \ldots, t_n,\ldots\}$ and let $\lambda_i \stackrel{\rm def}{=} P(T=t_i \mid T\geq t_i)$ . Write down $f(t_i) \stackrel{\rm def}{=} P(T=t_i), S(t_i)\stackrel{\rm def}{=}P(T>t_i)~(i=1,2,\ldots)$ in terms of $\lambda_j$ ‘s.

(解答)
For simiplicity, we denote $f_i \stackrel{\rm def}{=} f(t_i),~ S_i \stackrel{\rm def}{=} S(t_i),$ .

Note that

\lambda_i = \frac{f_i}{S_{i-1}}.

Thus, we have:

S_{i-1} = \frac{f_i}{\lambda_i}.

Furthermore, $S_{i-1}- S_i = f_i$ . So we have:

f_i = S_{i-1}-S_i = \frac{f_i}{\lambda_i} - \frac{f_{i+1}}{\lambda_{i+1}}.

By a simple rearrangement, we obtain:

\frac{f_{i+1}}{f_i} = \frac{\lambda_{i+1}}{\lambda_i} (\lambda_i - 1).

By considering $\prod_{i=1}^{n-1} (\cdot)$ , we have

\frac{f_{n}}{f_1} = \frac{\lambda_{n}}{\lambda_1} \prod_{i=1}^{n-1}(\lambda_i - 1).

Since $\lambda_1 = f_1$ , we have:

f_n = \lambda_n \prod_{i=1}^{n-1}(\lambda_i - 1).

And we also have:

S_n = \prod_{i=1}^{n}(\lambda_i - 1).

2. Let $X$ and $Y$ be mutually indepenent and (absolutely) continuous random variables with probability density funtions $f_X(x)$ and $f_Y(y)$ . Derive the probability density function of $Y$ conditioning $X-Y=0$ .

(方法) 首先考慮 $(Y, X-Y)$ 的聯合分布。有了聯合分布，我們就可以算條件分布。

(解答) Recall that a distribution of $(X,Y)$ is characterized by an expected value of $g(X,Y)$ where $g(\cdot)$ is an arbitrary bounded continuous function. We would like to first find the joint distributon of $(Y,X-Y)$ .

Let $W \stackrel{\rm def}{=} Y, \quad Z \stackrel{\rm def}{=} X-Y$ . Note that that $X = W+Z, \quad Y=W$ .

Consider the expected value of $g(W,Z)$ where $g$ is an arbitrary bounded continuous function.

E[g(W,Z)] = E[g(Y,X-Y)] = \int_{\mathbb{R}^2} g(y,x-y) f_X(x)f_Y(y)dxdy

By a variable transformation, the above integral is written by

E[g(W,Z)] \stackrel{\rm eq}{=} \int_{\mathbb{R}^2} g(w,z) f_X(w+z)f_Y(w)dwdz,

which implies that the joint probability density function of $(W,Z)$ is

f_{WZ}(w,z) \stackrel{\rm eq}{=} f_X(w+z)f_Y(w).

Therefore the marginal distribution of $Z$ is

f_{Z}(z) \stackrel{\rm eq}{=} \int_{\mathbb{R}}f_X(w+z)f_Y(w)dw.

The conditional probability density funcion $W \mid Z$ is

f_{W\mid Z}(w\mid z) \stackrel{\rm eq}{=} \frac{f_X(w+z)f_Y(w)}{\int_{\mathbb{R}}f_X(w+z)f_Y(w)dw}.

Finally, put $w \leftarrow y, z \leftarrow 0$ and we will obtain the desired distribution.

3.

In this problem, it is necessary to assuem that $R^2 \sim \chi^2_{(2)}$ to answer the 2nd question. So we assume that $R^2 \sim \chi^2_{(2)}$ in the 2nd part.

4.

5.