Question

Prove that the probability of at least one zero of \(Y(t)\) in the interval \(\left(t_{0}, t_{1}\right)\) is \((2 / \pi) \arccos \sqrt{t_{0} / t_{1}}\).

Short answer

To prove the given probability formula, we would ideally follow these steps: 1. Derive the probability density function for the zeros of \(Y(t)\). 2. Integrate the probability density function over the interval \((t_0, t_1)\) as \(\int_{t_0}^{t_1} f(t) dt\). 3. Calculate the probability and simplify the expression. 4. Compare the result to the given probability formula \(\frac{2}{\pi} \arccos\sqrt{\frac{t_0}{t_1}}\). However, since we don't have the functional form of \(Y(t)\), we cannot directly prove the statement. The mentioned process is the general method of solving such problems.

Step-by-step solution

Using the equation provided for \(Y(t)\), let's find the probability density function of the zeros of \(Y(t)\). To do this, we need to analyze the function and find any patterns regarding the distribution of its zeros. #Step 2: Integrate the Probability Density Function#

Once we have the probability density function, we need to integrate it over the interval \((t_0, t_1)\) to calculate the probability of at least one zero occurring within this interval. Therefore, we need to find the integral: \[\int_{t_0}^{t_1} f(t) dt\] #Step 3: Calculate the Probability and Simplify the Expression#

Now that we have the integral, we need to calculate the probability of at least one zero occurring in the interval \((t_0, t_1)\). We can do this by evaluating the integral and simplifying the expression. #Step 4: Compare the Result to the Given Probability Formula#

Finally, compare the result we got from the integral to the provided formula \(\frac{2}{\pi} \arccos\sqrt{\frac{t_0}{t_1}}\). If they are equal, then we have successfully proven the statement. Note: It is important to note that we cannot solve this problem completely as we don't have the functional form of \(Y(t)\). The process mentioned above is the general method of solving such problems.