$X \sim \mathcal{N}(1,\text{negligible variance})$ and $Y \sim \mathcal{N}(2,\text{negligible variance})$

\begin{equation*} Z= \begin{cases} X, & \ \text{w/pr}\quad p\\ Y, & \ \text{w/pr}\quad 1-p \\ \end{cases} \end{equation*}

What is the $\mathrm{E}[Z]$?


\begin{align*} \mathrm{E}[Z]&=p \times \mathrm{E}[X] +(1-p)\times \mathrm{E}[Y] \\ \mathrm{E}[Z]&=p \times 1+(1-p) \times 2 \\ \mathrm{E}[Z]&=p+2-2p \\ \mathrm{E}[Z]&=2-p \end{align*}

I have also used the transform method to prove the same answer. But the manual states that

"Take $X$ and $Y$ to be normal with means $1$ and $2$ respectively, and very small variances. Consider the random variable that takes the value of $X$ with some probability $p$ and the value of $Y$ with probability $1-p$. This random variable takes values near $1$ and $2$ with high probability, but takes values near its mean (which is $3-2p$) with relatively low probaility. Thus, this variable is not normal."

I can understand all other things but how he/she is saying mean is $3-2p$? Kindly someone guide, who is right and who is wrong?


tl;dnr version: The OP's final answer $E[Z] = 2-p$ is correct but the reasoning is not. The book's statement that $Z$ does not have a normal density is correct, but its mean computation incorrect (perhaps a typo).

As Chris Haug says, the OP's first statement (which the OP has deleted a few minutes ago) is incorrect. It is not true that $Z = pX + (1-p)Y$ regardless of whether $X$ and $Y$ are normal or not or independent or not. What is true is that $Z$ has what is called a mixture distribution. With $F$ denoting CDFs, the law of total probability says that

\begin{align}F_Z(\alpha) &\stackrel{\Delta}{=} P(Z\leq\alpha)\\&= pP(X\leq \alpha)+(1-p)P(Y\leq \alpha)\\&= pF_X(\alpha)+(1-p)F_Y(\alpha)\tag{1} \end{align}

and so, if $X$ and $Y$ are continuous random variables with density functions $f_X$ and $f_Y$ respectively, then $Z$ is also a continuous random variable with density function $$f_Z(\alpha) = pf_X(\alpha)+(1-p)f_Y(\alpha).\tag{2}$$ From $(2)$, it follows that \begin{align}E[Z] &= \int_{-\infty}^\infty \alpha f_Z(\alpha) \,\mathrm d\alpha\\ &= p \int_{-\infty}^\infty \alpha f_X(\alpha) \,\mathrm d\alpha + (1-p)\int_{-\infty}^\infty \alpha f_Y(\alpha) \,\mathrm d\alpha\\ &= pE[X] + (1-p)E[Y], \end{align} which for the OP's case works out to be $2-p$ as he computed. The OP's book's claim that $Z$ does not have a normal density even though $X$ and $Y$ have normal densities with different means is correct (as should be obvious from $(2)$ also) but the reasoning in support of this claim is dubious to say the least.

  • $\begingroup$ What is dubious about the reasoning? No Normal density can behave in the way described in the quotation. $\endgroup$
    – whuber
    Jun 20 '20 at 21:42

The very first line of your derivation is not correct (that's not the correct expression for the mixture $Z$). To illustrate, if $X$ and $Y$ are independent and if $p=0.7$, this is what these densities look like:


Note that the mixture $Z$ is bimodal, but $pX + (1-p)Y$ is actually a linear combination of independent normal variables, and is also normal.

However, what follows after that is indeed true: the mean is $2-p$, not $3-2p$.

The easiest way to see that $3-2p$ is incorrect is to plug in $p=0$, in which case $Z=Y$ and so the mean must be 2, but the manual's answer says 3. Perhaps in a previous version of the problem, $Y$ had mean 3.

  • $\begingroup$ Nowhere in the problem statement does it say that $X$ and $Y$ are independent. $\endgroup$ Jun 20 '20 at 19:58
  • $\begingroup$ @DilipSarwate That's correct, which is why I explicitly made that additional assumption in order to produce that illustration (it also doesn't say that $p=0.7$, but I had to pick something). The rest of what I said doesn't hinge on that assumption. $\endgroup$
    – Chris Haug
    Jun 20 '20 at 20:19
  • $\begingroup$ You don't need the assumption that $X$ and $Y$ are independent: the result that the density of $Z$ is a mixture of the densities of $X$ and $Y$ holds even when $X$ and $Y$ are dependent random variables and even when $Y=X+1$ (say). The variance does not need to be negligible either; it could even be infinite as with Cauchy random variables. $\endgroup$ Jun 20 '20 at 20:37
  • $\begingroup$ @DilipSarwate I am fully aware of that, as I just said. I wanted to show a visual representation of why that line was wrong, and that requires actually fixing a specific joint distribution to be able to draw it. Besides, a specific counterexample is legitimate proof that the statement is false, the fact that it doesn't need to be this specific example is completely irrelevant. $\endgroup$
    – Chris Haug
    Jun 20 '20 at 21:41

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