# Moment Generating Function of $\bar{M}−\bar{N}$

What is the Mean value of $$\bar{M}−\bar{N}$$; Moment Generating Function of $$\bar{M}−\bar{N}$$; and Variance of $$\bar{M}−\bar{N}.$$ Given $$M_1,M_2,\dots,M_n$$ is a random sample of size $$p$$ from the Gamma distribution $$(α,β)$$ and $$N_1,N_2 ,\dots,N_n$$ is a random sample of size $$q$$ from the distribution $$χ_r.$$ Where the random variable $$M$$ has the Gamma distribution $$(\alpha,\beta)$$ and the random variable $$N$$ has the Chi-Square distribution $$χ_r.$$ OK The random variable $$M$$ and the random variable $$N$$ are independent of each other. It can be assumed that $$\bar{M}$$ is the sample mean of the random variable $$M_1,M_2,\dots,M_n$$ and $$\bar{N}$$ is the sample mean of the random variable $$N_1,N_2,\dots,N_n$$.

My Attempt: (Anyone can help me for my answer, is it right or not? Thank you)

Mean value of $$\bar{M}-\bar{N}$$

The mean of the difference of two independent random variables is the difference of their means. Since $$\bar{M}$$ and $$\bar{N}$$ are the sample means of independent random samples, they are independent of each other. Therefore, the mean of $$\bar{M}-\bar{N}$$ is the difference of their means, which is:

$$E(\bar{M}-\bar{N}) = E(\bar{M}) - E(\bar{N})$$

The mean of a sample mean is equal to the population mean. The population mean of the Gamma distribution $$(\alpha,\beta)$$ is $$\frac{\alpha}{\beta}$$, and the population mean of the Chi-Square distribution $$\chi_r$$ is $$r$$. Therefore, the mean of $$\bar{M}-\bar{N}$$ is:

$$E(\bar{M}-\bar{N}) = \frac{\alpha}{\beta} - r$$

Moment Generating Function of $$\bar{M}-\bar{N}$$

The moment generating function (MGF) of a random variable $$X$$ is defined as:

$$M_X(t) = E\left[e^{tX}\right]$$

The MGF of the sum of two independent random variables is the product of their MGFs. Therefore, the MGF of $$\bar{M}-\bar{N}$$ is:

$$M_{\bar{M}-\bar{N}}(t) = M_{\bar{M}}(t) M_{\bar{N}}(-t)$$

The MGF of the sample mean of a random sample of size $$n$$ from a distribution with MGF $$M_X(t)$$ is:

$$M_{\bar{X}}(t) = \left[ M_X\left(\frac{t}{n}\right) \right]^n$$

Therefore, the MGF of $$\bar{M}-\bar{N}$$ is:

$$M_{\bar{M}-\bar{N}}(t) = \left[ M_M\left(\frac{t}{p}\right) \right]^p \left[ M_N\left(-\frac{t}{q}\right) \right]^q$$

Variance of $$\bar{M}-\bar{N}$$

The variance of the difference of two independent random variables is the sum of their variances. Since $$\bar{M}$$ and $$\bar{N}$$ are independent random samples, their variances are additive. Therefore, the variance of $$\bar{M}-\bar{N}$$ is:

$$Var(\bar{M}-\bar{N}) = Var(\bar{M}) + Var(\bar{N})$$

The variance of a sample mean is equal to the population variance divided by the sample size. The population variance of the Gamma distribution $$(\alpha,\beta)$$ is $$\frac{\alpha^2}{\beta^2}$$, and the population variance of the Chi-Square distribution $$\chi_r$$ is $$2r$$. Therefore, the variance of $$\bar{M}-\bar{N}$$ is:

$$Var(\bar{M}-\bar{N}) = \frac{\alpha^2}{p\beta^2} + \frac{2r}{q}$$

I hope this helps!