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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!

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