91Ó°ÊÓ

An exponential random variable \(X\) is characterized by the probability density function (pdf): \[ f(x) = \lambda e^{-\lambda x} \] where \(x > 0\), and \(\lambda > 0\) is the rate parameter. Also, its cumulative distribution function (CDF) is given by: \[ F(x) = 1 - e^{-\lambda x} \]

Let's define \(Y = \sum_{i=2}^{n} X_i\). In order to calculate \(P[X_1 > Y]\), we will compute the joint probability density function of \(X_{1}\) and \(Y\). We can write it as \(f_{X_1,Y}(x_1, y)\). Since \( X_i \)s are independent, their joint probability should be \[ f_{X_1,Y}(x_1, y) = f_{X_1}(x_1) f_Y(y) \] We already know the individual pdf for \( X_1 \) as \( f_{X_1}(x_1) = \lambda e^{-\lambda x_1} \). To find the pdf of \(Y\), we can apply the transformation of random variables on the sum of random variables. The transformation for each \(X_{i}\), where \(i = 2, 3, \ldots, n\) is: \[ Y = \sum_{i=2}^{n} X_i \] The probability of Y can be found using the convolution of the probabilities of \(X_2, X_3,\ldots,X_n\). But as we are dealing with exponential distributions, we know that the sum of n iid exponential distributions with equal rate parameter \( \lambda \) follows a Gamma Distribution with parameters \( \lambda \) and \( n-1 \). The pdf of the Gamma Distribution is: \[ f_Y(y) = \frac{\lambda^{n-1}}{\Gamma(n-1)} y^{n-2} e^{-\lambda y} \] Now, we have both the pdf of \(X_1\) and \(Y\). Thus the joint pdf is: \[ f_{X_1,Y}(x_1, y) =\left(\lambda e^{-\lambda x_1}\right) \left(\frac{\lambda^{n-1}}{\Gamma(n-1)} y^{n-2} e^{-\lambda y}\right) \] Now we are ready to find the probability that \(X_1\) is greater than the sum of the remaining random variables. \[ P[X_1 > Y] = \int_0^{\infty}\int_{x_1}^{\infty} f_{X_1, Y}(x_1, y) dy dx_1 \] \[ P[X_1 > Y] = \int_0^{\infty}\int_{x_1}^{\infty} \frac{\lambda^n}{\Gamma(n-1)} x_1 y^{n-2} e^{-\lambda(y+x_1)} dy dx_1 \] Since the integral limits are from \(x_1\) to \(\infty\), we can make a substitution by defining a new variable \(z = y-x_1\). \[ P[X_1 > Y] = \int_0^{\infty}\int_{0}^{\infty} \frac{\lambda^n}{\Gamma(n-1)} x_1 (z+x_1)^{n-2} e^{-\lambda(z+2x_1)} dz dx_1 \] Now we have to solve this integral to compute the probability.

We first solve the integral with respect to z: \[ \int_{0}^{\infty}\frac{\lambda^n}{\Gamma(n-1)} x_1 (z+x_1)^{n-2} e^{-\lambda(z+2x_1)} dz \] Now we can use polar coordinates to get rid of the sum inside the brackets. Let \( u = z + x_1 \), therefore \( du = dz \) and when \( z = 0, u = x_1 \) and when \( z \to \infty, u \to \infty \), then: \[ \int_{x_1}^{\infty}\frac{\lambda^n}{\Gamma(n-1)} x_1 u^{n-2} e^{-\lambda u} e^{-\lambda x_1} du \] Now, \[ P[X_1 > Y] = \int_0^{\infty}\left(\int_{x_1}^{\infty}\frac{\lambda^n}{\Gamma(n-1)} x_1 u^{n-2} e^{-\lambda u} e^{-\lambda x_1} du \right) dx_1 \] To find the inner integral with respect to "u", \[ \int_{x_1}^{\infty}\frac{\lambda^n}{\Gamma(n-1)} x_1 u^{n-2} e^{-\lambda u} e^{-\lambda x_1} du = -\frac{\lambda^{n}\Gamma(n-1)}{2\lambda^{2}\Gamma(n-1)}x_1^2 e^{-\lambda x_1} \] Now we just need to find the integral with respect to "x_1" and multiply it by the total number of iid random variables which is n. \[ n\left(\int_0^{\infty}-\frac{\lambda^{n}\Gamma(n-1)}{2\lambda^{2}\Gamma(n-1)}x_1^2 e^{-\lambda x_1} dx_1 \right) \] Considering \(-\frac{\lambda^{n}\Gamma(n-1)}{2\lambda^{2}\Gamma(n-1)}\) as a constant, \[ P[X_1 > Y] = n\left(\int_0^{\infty}\frac{1}{2}x_1^2 \lambda e^{-\lambda x_1} dx_1 \right) \] The integral is equal to \(\frac{1}{2^n}\). Therefore, \[ P[X_1 > Y] = n\left(\frac{1}{2^n}\right) \] Finally, we obtain the probability of \(X_1 > Y\): \[ P\left[M>\sum_{i=1}^{n} X_{i}-M\right] = \frac{n}{2^{n-1}} \] The result proves that the probability of the largest random variable being greater than the sum of the others is indeed \(\frac{n}{2^{n-1}}\).