# Finding a critical region for a mixture

Let $$\ X_1 , X_2$$ be two iid random variables with normal N ( $$\theta,1$$) distribution. Further , consider bernoulli random variable V with P(V=1) = $$\frac{1}{4}$$ and which is independent of $$\ X_1 , X_2$$ .

Define $$\ X_3$$ as = $$\ X_1$$, if V=0 $$\ X_2$$, if V=1

For testing hypothesis H0 : $$\theta =0$$ versus H1 : $$\theta =1$$

Reject H0 if $$\frac{\ X_1 + X_2 +X_3 }{3}$$ >C

Find C such that test size becomes 0.05 .

Now in this question I could understand only two things , one that V is an indicator variable and bernoulli distribution and second that it should be converted to Z before adding two critical region.

My attempt

$$\ X_3 = \ X_1*P(v=0) + \ X_2*P(v=1 )$$

$$E(\ X_3) = E(\ X_1*P(v=0)) + E(\ X_2*P(v=1 ))$$

$$E(\ X_3) = (\theta *3/4) + (\theta*1/4)$$

$$E(\ X_3) = \theta$$

Now $$Var (\ X_3) = Var(\ X_1 *3/4) + var(\ X_2*1/4)$$

$$Var (\ X_3) = \frac{9}{16} + \frac{1}{16}$$

$$Var (\ X_3) = \frac{10}{16}$$

$$\frac{\ X_1 + X_2 +X_3 }{1} \ ~~~ N( 3\theta , ??)$$

$$\frac{\ X_1 + X_2 +X_3 }{3} \ ~~~ N( \theta , ??)$$ Please help to proceed .

• This exercise asks you to find an upper quantile of $Y=(X_1+X_2+X_3)/3.$ One straightforward approach, then, would begin by finding an expression for the CDF of $Y.$ What do you obtain?
– whuber
Jul 12 at 17:57
• @simran You're mistaken; $X_3$ is not an indicator, nor does it have a Bernoulli distribution. $V$ is an indicator. Jul 13 at 2:44
• @glen_b then what would be the distribution of X3 , can you please suggest a book on inference inference that could help to tackle such questions , and others also that I have posted , if you could visit my profile , please Jul 13 at 3:37
• I commented so that you would clarify/correct details in your question. Please do so. Jul 13 at 7:03
• Sometimes one can succeed in course work by guessing a formula or result -- but in the real world, that is rarely effective. Thus, what is most important is to explain why you think the sum of the $X_i$ might have a particular distribution: how did you derive it?
– whuber
Jul 13 at 14:36

A mixture of identically distributed variables has the same distribution -- but is not independent of its component variables.

Some of the information in this exercise is just a distraction. Stripped to its essential underlying idea, the problem is the following:

Let $$X_1,X_2$$ be independent identically distributed random variables. Use an independent Bernoulli$$(p)$$ variable $$V$$ to define $$X_3 = X_1$$ when $$V=0$$ and otherwise $$X_3=X_2.$$ What is the distribution of $$Y = X_1+X_2+X_3$$?

The iid assumption implies $$(X_1,X_2)$$ and $$(X_2,X_1)$$ have the same distribution: that is, we may switch the roles of the $$X_i.$$ Upon doing this with the random variable $$2X_1+X_2$$ it is immediate that

$$2X_1+X_2$$ and $$X_1 + 2X_2$$ have the same distribution.

Let this common distribution function be $$F.$$ This means only that for any number $$y,$$

$$\Pr(2X_1+X_2 \le y) = F(y) = \Pr(X_1+2X_2 \le y.)$$

The distribution function of $$Y$$ is found by studying the two possibilities for $$V,$$ using the facts that when $$V=0,$$ $$Y=2X_1+X_2$$ and when $$V=1,$$ $$Y=X_1+2X_2.$$ The law of total probability asserts

\begin{aligned} \Pr(Y\le y) &= \Pr(V=0)\Pr(2X_1 + X_2 \le y) + \Pr(V=1)\Pr(X_1+2X_2\le y)\\ & = (1-p)F(y) + pF(y) \\ &= F(y). \end{aligned}

Thus,

$$Y$$ has the same distribution as $$2X_1+X_2$$ and $$X_1 + 2X_2.$$

The rest is mopping up: in your circumstance, where the $$X_i$$ have Normal distributions and are independent, $$Y$$ must therefore have a Normal distribution and its parameters are easily computed. From that it's simple to find the value of $$C$$ in the question.

The joint distribution of all the relevant variables is unusual and instructive. In this scatterplot matrix of 2000 realizations (with $$\theta=5$$) I have colored the points where $$V=0$$ in blue and those where $$V=1$$ in red. Only $$X_1$$ and $$X_2$$ are independent, as indicated by their circular-cloud scatterplots. The scatterplots with $$Y$$ include some singular parts on the diagonal, reflecting the fact that often $$Y$$ equals $$2X_1+X_2$$ or $$X_1+2X_2.$$

I think that,

$$X_3 = \cases{X_1, V = 0\\X_2, V = 1}$$

Therefore,

$$X_3 \sim N(\theta, 1)$$

Check if $$X_3$$ is independent of $$X_1$$ and $$X_2$$.

So what you have presented is the average of 3 iid observations of the variable $$X \sim N(\theta, 1)$$.

It is known that $$\bar{x} \sim N(\theta, \frac{\sigma}{n})$$. Which in this case $$\bar{x} \sim N(\theta, \frac{1}{\sqrt{3}})$$.

So I reject $$H_0$$ if $$P(\bar{x} > c) = P(Z > \frac{c - \theta }{1/\sqrt{3}}) = P(Z > \sqrt{3}(c - \theta)) = 0.05$$.

It follows then that

$$c - \theta = \frac{1.64}{\sqrt{3}} \Rightarrow c = \theta + \frac{1.64}{\sqrt{3}}$$

• $X_3$ obviously is not independent of $X_1$ and $X_2:$ with 100% probability, it is equal to one or the other of them!
– whuber
Jul 14 at 18:24
• It makes sense @whuber. I did a simulation in R and noticed I got a wrong root. I have to think more about this problem. Jul 14 at 18:55
• A scatterplot matrix is informative. In R: n <- 1e3; theta <- 5; v <- runif(n) < 3/4; x.1 <- rnorm(n, theta); x.2 <- rnorm(n, theta); x.3 <- ifelse(v, x.2, x.1); pairs(cbind(x.1, x.2, x.3), col=ifelse(v, "red", "blue"))
– whuber
Jul 14 at 19:39