# Normally distributed conditional random variables

The weight of the boys is distributed normally with mean $$3.4$$ and standard deviation $$0.3$$, and the weight of the girls is distributed normally with mean $$3.2$$ and standard deviation $$0.3$$. If the ratio of the boys and girls is $$50:50$$, what is the standard deviation of the weight of a baby who is randomly chosen?

My approach is the following:

Let $$X$$ be the weight of a baby, $$B$$ be the boys and $$G$$ be the girls. Then $$X\mid B\sim N(3.4,0.3^2),\ X\mid G\sim N(3.2,0.3^2).$$ What I need to find is $$\sigma_X.$$

...And at this point I realized that the definitions of $$B,G$$ are vague. How can I convert this problem into a computable mathematical language?

EDIT:

Let $$G$$ be a random variable that takes only $$2$$ values: $$M$$ for male and $$F$$ for female. And each has the probability $$1\over 2$$. That is,

$$P(G=M)={1\over 2},\ P(G=F)={1\over 2}.$$

And if $$X$$ is the random variable of the weight of a baby, we have

$$(X\mid G=M)\sim N(3.4,0.3^2),\ (X\mid G=F)\sim N(3.2,0.3^2).$$

And what I have to find is $$Var(X).$$

By the law of total variance,

\begin{align}Var(X)&=E(Var(X\mid G))+Var(E(X\mid G))\\ &=E[E(X^2\mid G)-E(X\mid G)^2] + [E(E(X\mid G)^2)-E(E(X\mid G))^2]\\ &=E(E(X^2\mid G))-E(E(X\mid G)^2) + E(E(X\mid G)^2)-E(E(X\mid G))^2\\ &=E(E(X^2\mid G))-E(E(X\mid G))^2.\end{align}

And since $$E(X\mid G)$$ is a random variable depending only on $$G$$,

\begin{align}E(E(X\mid G)) &=P(G=M)\cdot E(X\mid G=M)+P(G=F)\cdot E(X\mid G=F)\\ &= {1\over 2}\cdot 3.4+{1\over 2}\cdot 3.2\\ &= 3.3,\end{align}

\begin{align}E(E(X^2\mid G)) &=P(G=M)\cdot E(X^2\mid G=M)+P(G=F)\cdot E(X^2\mid G=F) \\ &={1\over 2}\cdot [Var(X\mid G=M)+E(X\mid G=M)^2]+{1\over 2}\cdot [Var(X\mid G=F)+E(X\mid G=F)^2]\\ &={1\over 2}\cdot [0.3^2+3.4^2]+{1\over 2}\cdot [0.3^2+3.2^2]\\ &=10.99 \end{align}

Hence

$$Var(X)=10.99-3.3^2=10.99-10.89=0.1$$

And

$$\sigma_X=\sqrt{Var(X)}=0.316227766$$

• this is like a mixture model $f(x)=.5 N(3.4,.09)+.5N(3.2,.09)$. is it? Mar 27 '19 at 1:42
• @masoud Right, then is it correct to say $f(x)\sim N(\frac{3.4+3.2}{2},\frac{0.09+0.09}{2})=N(3.3,0.09)$? Mar 27 '19 at 1:55
• To calculate sd you could use the law of total variance. Mar 27 '19 at 2:29
• In general your model be like $p n(a,b) +qn(c,d)$(check identifability) and estimate all parameters include p. even p can be random variable. that in your case the values be .5 . Mar 27 '19 at 2:35
• You could but you should not. (Could means you could have a bad analysis) It is a bad option. .This ia a mixture model. The target distribution is mixes of two distributions. For example it has two mods and difference from a unimodel normal. you lose the information that may give u more better estimation. If you do that, you assumes that every observation have a same mean and variance that is not correct(and same distribution). Mar 27 '19 at 2:47

Let $$(X\mid G=m)\sim\mathcal N(\mu_m,\sigma_m^2)$$ and $$(X\mid G=f)\sim\mathcal N(\mu_f,\sigma_f^2)$$ and $$G\sim\mathcal U\{m,f\}$$
\begin{align}\mathsf E(X)&=\mathsf E\mathsf E (X\mid G)\\&=\tfrac 12(\mu_m+\mu_f)\\&=\tfrac 12(3.4+3.2)\\&=3.3\\[3ex]\mathsf {Var}(X)&=\mathsf{Var}\mathsf E(X\mid G)+\mathsf E\mathsf{Var}(X\mid G)\\&= \mathsf E\mathsf E^2(X\mid G)-\mathsf E^2\mathsf E(X\mid G)+\mathsf E\mathsf{Var}(X\mid G)\\&=\tfrac 12(\mu_m^2+\mu_f^2)-(\tfrac 12(\mu_m+\mu_f))^2+\tfrac 12(\sigma_m^2+\sigma_f^2)\\&=\tfrac 12(3.4^2+3.2^2)-(\tfrac 12(3.4+3.2))^2+0.3^2\\&= 10.90-10.89+0.09\\&=0.10\end{align}