Recall that an exponential random variable 𝑋~exp (𝜆) has mean 𝜆 1 and pdf given by: 𝑓(𝑥) = 𝜆𝑒 −𝜆𝑥 on 𝑥 ≥ 0. (b) Suppose that 𝑋1 and 𝑋2 are independent exponential random variables with mean 𝜆 1 . Let 𝑇 = min (𝑋1 ,𝑋2 ). Find the cdf of 𝑇. (Hint: what is 𝑃(𝑇 ≥ 𝑡)?)教我怎么做

首先，我们可以使用下面的公式计算 $P(T \geq t)$：

$$P(T \geq t) = P(X_1 \geq t, X_2 \geq t)$$

因为 $X_1$ 和 $X_2$ 是独立的，所以我们可以将概率分解为：

$$P(X_1 \geq t, X_2 \geq t) = P(X_1 \geq t) \cdot P(X_2 \geq t)$$

由于 $X_1$ 和 $X_2$ 都是指数分布，所以：

$$P(X_i \geq t) = e^{-\lambda_i t}, \ i = 1,2$$

因此，我们有：

$$P(T \geq t) = P(X_1 \geq t) \cdot P(X_2 \geq t) = e^{-\lambda_1 t} \cdot e^{-\lambda_2 t} = e^{-(\lambda_1 + \lambda_2) t}$$

现在我们可以使用 $P(T \geq t)$ 来计算 $T$ 的累积分布函数 (CDF)：

$$F_T(t) = P(T \leq t) = 1 - P(T \geq t) = 1 - e^{-(\lambda_1 + \lambda_2) t}$$

因此，$T$ 的 CDF 为 $F_T(t) = 1 - e^{-(\lambda_1 + \lambda_2) t}$。

相关问题

Recall that an exponential random variable 𝑋~exp (𝜆) has mean 𝜆 1 and pdf given by: 𝑓(𝑥) = 𝜆𝑒 −𝜆𝑥 on 𝑥 ≥ 0. (a) Compute 𝑃(𝑋 ≥ 𝑥). (b) Suppose that 𝑋1 and 𝑋2 are independent exponential random variables with mean 𝜆 1 . Let 𝑇 = min (𝑋1 ,𝑋2 ). Find the cdf of 𝑇. (Hint: what is 𝑃(𝑇 ≥ 𝑡)?)

(a) 𝑃(𝑋 ≥ 𝑥) = ∫𝑥∞ 𝑓(𝑥) d𝑥 = ∫𝑥∞ 𝜆𝑒^−𝜆𝑥 d𝑥 = e^−𝜆𝑥

(b) 𝑃(𝑇 ≥ 𝑡) is the probability that both 𝑋1 and 𝑋2 are greater than or equal to 𝑡, which is given by:

𝑃(𝑇 ≥ 𝑡) = 𝑃(𝑋1 ≥ 𝑡)𝑃(𝑋2 ≥ 𝑡) = e^−𝜆1𝑡 * e^−𝜆1𝑡 = e^(−2𝜆1𝑡)

To find the cdf of 𝑇, we can use the fact that the cdf is the integral of the pdf:

𝐹𝑇(𝑡) = 𝑃(𝑇 ≤ 𝑡) = 1 − 𝑃(𝑇 ≥ 𝑡) = 1 − e^(−2𝜆1𝑡)

Therefore, the cdf of 𝑇 is 𝐹𝑇(𝑡) = 1 − e^(−2𝜆1𝑡).

# 实现高斯核函数 def rbf_kernel(x1, x2): sigma=1.0 return np.exp(-np.linalg.norm(x1-x2,2)**2/sigma) # 加载葡萄酒识别数据集 wine = datasets.load_wine() # 处理数据和标签 X= wine["data"][:,(0,1)] y = 2 * (wine["target"]==1).astype(np.int64).reshape(-1,1) - 1 # 按照一定比例划分训练集和测试集（测试集占0.4） X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.4, random_state=5) # 定义KernelSVM模型 # 训练模型 # 用Pyplot作图工具绘制模型预测边界 x0s = np.linspace(10, 15, 100) x1s = np.linspace(0, 7, 100) x0, x1 = np.meshgrid(x0s, x1s) W = np.c_[x0.ravel(), x1.ravel()] u= model.predict(W).reshape(x0.shape) plt.plot(X_train[:, 0][y_train[:,0]==1] , X_train[:, 1][y_train[:,0]==1], "gs") plt.plot(X_train[:, 0][y_train[:,0]==-1], X_train[:, 1][y_train[:,0]==-1], "ro") plt.contourf(x0, x1, u, alpha=0.2) plt.show() #计算指标 from sklearn.metrics import accuracy_score from sklearn.metrics import recall_score from sklearn.metrics import precision_score from sklearn.metrics import f1_score accuracy = accuracy_score(y_test, y_pred) precision = precision_score(y_test, y_pred) recall = recall_score(y_test, y_pred) f1 = f1_score(y_test, y_pred) print("accuracy = {}".format(accuracy)) print("precision = {}".format(precision)) print("recall = {}".format(recall)) print("f1 = {}".format(f1))补全代码

抱歉，上面的代码仍然不完整，缺少了定义KernelSVM模型和训练模型的部分。以下是完整的代码：

# 实现高斯核函数 
def rbf_kernel(x1, x2):     
    sigma=1.0     
    return np.exp(-np.linalg.norm(x1-x2,2)**2/sigma)           

# 加载葡萄酒识别数据集   
wine = datasets.load_wine()  

# 处理数据和标签 
X= wine["data"][:,(0,1)]
y = 2 * (wine["target"]==1).astype(np.int64).reshape(-1,1) - 1  

# 按照一定比例划分训练集和测试集（测试集占0.4） 
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.4, random_state=5)  

# 定义KernelSVM模型  
svm_clf = Pipeline([
        ("scaler", StandardScaler()),
        ("svm_clf", SVC(kernel="rbf", gamma=5, C=0.001))
    ])

# 训练模型  
svm_clf.fit(X_train, y_train.ravel())

# 用Pyplot作图工具绘制模型预测边界 
x0s = np.linspace(10, 15, 100) 
x1s = np.linspace(0, 7, 100) 
x0, x1 = np.meshgrid(x0s, x1s) 
W = np.c_[x0.ravel(), x1.ravel()] 
u= svm_clf.predict(W).reshape(x0.shape) 
plt.plot(X_train[:, 0][y_train[:,0]==1] , X_train[:, 1][y_train[:,0]==1], "gs") 
plt.plot(X_train[:, 0][y_train[:,0]==-1], X_train[:, 1][y_train[:,0]==-1], "ro") 
plt.contourf(x0, x1, u, alpha=0.2) 
plt.show() 

# 计算指标 
y_pred = svm_clf.predict(X_test)
accuracy = accuracy_score(y_test, y_pred) 
precision = precision_score(y_test, y_pred) 
recall = recall_score(y_test, y_pred) 
f1 = f1_score(y_test, y_pred) 

print("accuracy = {}".format(accuracy)) 
print("precision = {}".format(precision)) 
print("recall = {}".format(recall)) 
print("f1 = {}".format(f1))

其中，`SVC`是Scikit-Learn库中的SVM模型，`Pipeline`是Scikit-Learn库中的管道模型，用于将多个数据预处理和模型组合成一个整体。