We are given that $X_1, X_2, \dots, X_n$ are independent random variables, each following an exponential distribution with rate parameter $\lambda$, denoted $E(\lambda)$. We need to show that the random variable

$T = 2\lambda \sum_{i=1}^n X_i$

has a chi-squared distribution with $2n$ degrees of freedom, i.e., $T \sim \chi^2_{2n}$.

Step 1: Understand the distribution of $X_i$

Each $X_i$ is exponentially distributed with parameter $\lambda$. The probability density function (PDF) of $X_i$ is:

$f_{X_i}(x) = \lambda e^{-\lambda x}, \quad x \geq 0.$

Step 2: Consider the transformed variable $Y_i = 2\lambda X_i$

We need to show that $2\lambda X_i$ follows a Gamma distribution with shape $1$ and rate $\frac{1}{2}$, i.e., $\Gamma(1, \frac{1}{2})$, which is equivalent to a $\chi^2_2$ distribution. To do this, we first compute the PDF of $Y_i = 2\lambda X_i$.

Let’s find the distribution of $Y_i$. The CDF of $Y_i$ is:

$P(Y_i \leq y) = P(2\lambda X_i \leq y) = P\left( X_i \leq \frac{y}{2\lambda} \right).$

Since $X_i$ is exponentially distributed, its CDF is:

$F_{X_i}(x) = 1 - e^{-\lambda x}.$

Substituting $x = \frac{y}{2\lambda}$, we get:

$F_{Y_i}(y) = 1 - e^{-\frac{y}{2}}.$

Now, differentiate this with respect to $y$ to find the PDF of $Y_i$:

$f_{Y_i}(y) = \frac{d}{dy} \left( 1 - e^{-\frac{y}{2}} \right) = \frac{1}{2} e^{-\frac{y}{2}}, \quad y \geq 0.$

This is the PDF of a $\Gamma(1, \frac{1}{2})$ distribution, which is also equivalent to a $\chi^2_2$ distribution.

Step 3: Sum of independent $\chi^2_2$ variables

Since $Y_1, Y_2, \dots, Y_n$ are independent and each follows a $\chi^2_2$ distribution (because each $2\lambda X_i \sim \Gamma(1, \frac{1}{2})$), the sum of these independent variables:

$T = \sum_{i=1}^n Y_i = 2\lambda \sum_{i=1}^n X_i$

follows a chi-squared distribution with $2n$ degrees of freedom. Specifically:

$T \sim \chi^2_{2n}.$

Conclusion

We have shown that if $X_1, X_2, \dots, X_n$ are independent $E(\lambda)$ random variables, then

$T = 2\lambda \sum_{i=1}^n X_i \sim \chi^2_{2n}.$

Would you like further clarification or details on any part of this solution?

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Here are some related questions for you to explore further:

1. What is the general relationship between a Gamma distribution and the chi-squared distribution?
2. How do we derive the distribution of the sum of independent random variables?
3. Can you prove the moment-generating function of the chi-squared distribution?
4. How does the shape and scale parameter of the Gamma distribution affect the chi-squared distribution?
5. What happens if we use a different scaling factor, such as $3\lambda$ instead of $2\lambda$?

Tip: The chi-squared distribution is a special case of the Gamma distribution with integer shape parameters. It is often useful in statistical inference, particularly in hypothesis testing and confidence interval estimation.