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I want to add two variance. Example: $\sigma_1^2=10$ and $\sigma_2^2=30.$ Will $\sigma_{total}^2 = \sigma_1^2+\sigma_2^2 =40?$

Can I add the two variance directly or is there some law to add two variances.

Please help

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    Firstly, variances cannot be negative, you've got something wrong there. Secondly do you know if the random variables involved are independent or not? $\mathrm{Var}[X_1 + X_2] = \mathrm{Var}[X_1] + \mathrm{Var}[V_2]$ if they're independent, or uncorrelated, but without that information you cannot just add them and get a meaningful result. – 2017-02-25
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    thank you for your reply. If they are independent, then i can directly add the variance. – 2017-02-25
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    Yes, add as @stochasticboy says, provided the random variables are independent. // BTW, in your Question it seems you had some negative variances. There's no such thing. So I changed the question a bit, when I put it into MathJaX notation for readability. If you think I did damage, please change it again, and explain what you mean. – 2017-02-25
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    No problem sir...now i just have to make sure that my random variables are independent. With the total variance, I am going to generate random variables using normal distribution (mean=0,total_variance) – 2017-02-25

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Here is a quick demo that independence is crucial. Let $X$ be a random variable and let $Y$ be exactly the same random variable: $X \equiv Y.$ (Not just a random variable with the same distribution.) Certainly, then $X$ and $Y$ are associated, not independent. Let $V(X) = V(Y) = \sigma^2.$

If we could ignore independence, then we would have equality throughout in the following relationship, and thus be able to prove that $2 = 4.$

$$ 2\sigma^2 = V(X) + V(Y) \ne V(X + Y) = V(2X) = 4V(X) = 4\sigma^2.$$

Note: Here are some important formulas about expectations and variances of random variables. You should find these (or similar) formulas in your book and look at any examples given there.

$E(a + bX + cY) = a + bE(X) + cE(Y),$ regardless of independence. In particular, setting $c=0,$ we have $E(a + bX) = a + bE(X),$

$V(a + bX + cY) = b^2V(X) + c^2V(Y),$ only if $X$ and $Y$ are uncorrelated. In particular, $V(a + bX) = b^2V(X).$ [Above I used this with $a = 0,\, b=2.$]

Also, setting $a=0,\, b=1,\, c = -1,$ we have $V(X - Y) = V(X) + V(Y),$ only if $X$ and $Y$ are uncorrelated.