# Discrete-continuous random variable!

#### OhMyMarkov

##### Member
Hello everyone!

I'm looking at the following random variables:

$Z_1$ is normally distributed with zero mean and variance $\sigma _1 ^2$
$Z_2$ is normally distributed with zero mean and variance $\sigma _2 ^2$

$B$ is Bernoulli with probability of success $p$.

$X$ is a random variable that takes $Z_1$ if $B=1$ and $Z_2$ if $B=0$.

What is the variance of $X$?

#### chisigma

##### Well-known member
Re: Dicrete-continuous random variable!

Hello everyone!

I'm looking at the following random variables:

$Z_1$ is normally distributed with zero mean and variance $\sigma _1 ^2$
$Z_2$ is normally distributed with zero mean and variance $\sigma _2 ^2$

$B$ is Bernoulli with probability of success $p$.

$X$ is a random variable that takes $Z_1$ if $B=1$ and $Z_2$ if $B=0$.

What is the variance of $X$?
The 'instinctive' answer should be...

$\displaystyle \sigma_{X}^{2}= p\ \sigma_{1}^{2} + (1-p)\ \sigma_{2}^{2}$ (1)

Kind regards

$\chi$ $\sigma$

#### Bacterius

##### Well-known member
MHB Math Helper
Re: Dicrete-continuous random variable!

My statistics are a bit rusty, but we have:

$$P(B = 1) = p$$
$$P(B = 0) = 1 - p$$

And we're given that:

$$VAR(X | B = 1) = \sigma_1^2$$
$$VAR(X | B = 0) = \sigma_2^2$$

And since $Z_1$ and $Z_2$ are independent, you can just add the variances up:

$$VAR(X) = P(B = 1) VAR(X | B = 1) + P(B = 0) VAR(X | B = 0)$$

Which gives:

$$VAR(X) = p \sigma_1^2 + (1 - p) \sigma_2^2 = \sigma_1^2 + (\sigma_2^2 - \sigma_1^2) p$$

I've checked the result empirically.

#### awkward

##### Member
Re: Dicrete-continuous random variable!

[snip]

And since $Z_1$ and $Z_2$ are independent, you can just add the variances up:

$$VAR(X) = P(B = 1) VAR(X | B = 1) + P(B = 0) VAR(X | B = 0)$$

[snip]
Hi Bacterius,

I don't think that is true, in general. Have you thought about what would change if $Z_1$ and $Z_2$ did not have the same mean?

$$var[X] = E[X^2] - E[X]^2$$

#### Bacterius

##### Well-known member
MHB Math Helper
Re: Dicrete-continuous random variable!

Hi Bacterius,

I don't think that is true, in general. Have you thought about what would change if $Z_1$ and $Z_2$ did not have the same mean?

$$var[X] = E[X^2] - E[X]^2$$
Yes, it only works in this particular case. I did not consider beyond the problem asked.

#### rashtastic

##### New member
Re: Dicrete-continuous random variable!

Cool problem.

Assign $Z_{i}$ to have mean $\mu_{i}$ and pdf $f_{i}(x)$.

Everything should follow if we find a pdf $f(x)$ for $X$.

$f(x)=P(X=x)=P(X=x|Z_{1})P(Z_{1})+P(X=x|Z_{2})P(Z_{2})$
(Law of Total Probability)
$=pf_{1}(x)+(1-p)f_{2}(x).$

$E(X)=\int_{\mathbb{R}}x*f(x)dx$
$=\int_{\mathbb{R}}{x*[pf_{1}(x)+(1-p)f_{2}(x)]}dx$
$=p\int_{\mathbb{R}}{x*f_{1}(x)dx} + (1-p)\int_{\mathbb{R}}{{x}*f_{2}dx}$
$=pE(Z_{1})+(1-p)E(Z_{2})$
$=p\mu_{1}+(1-p)\mu_{2}$.

$E(X^{2})=\int_{\mathbb{R}}{x^{2}}f(x)dx$
$=\int_{\mathbb{R}}{x^{2}}[pf_{1}(x)+(1-p)f_{2}(x)]dx$
$=p\int_{\mathbb{R}}{x^{2}f_{1}(x)dx}+(1-p)\int_{\mathbb{R}}{x^{2}f_{2}(x)dx}$
$=pE(Z_{1}^{2})+(1-p)E(Z_{2}^{2})$
$=p(\sigma_{1}^{2}+\mu_{1}^2)+(1-p)(\sigma_{2}^{2}+\mu_{2}^2)$
(Because we know $var(Z)=E(Z^{2})-E(Z)^{2}$)

$var(X)=E(X^{2})-E(X)^2$, for which I am getting
$p\sigma_{1}^{2}+(1-p)\sigma_{2}^{2}+p(1-p)(\mu_{1}-\mu_{2})^{2}$.

This concurs with past answers, but it's strange. The problem is what the variance of X would be if the means are very different like two distributions:

..... .. . .. ... ... _________________________ . .. . ... . ... .. .... ..

Calculating the total variance should involve the distance from each point to the total mean where the total mean is pretty far from each distribution's mean. But if the above work is correct, all the information you need about how the total variance is changed when the means are changed is in the difference between the means. (Not the sum- made a mistake typing it up).

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