Is RSA key size the size of private key exponent?
No, the RSA key size is not the size of the private key exponent. It is customarily the number of bits in the public modulus (which is known as $N$). In other words, the key size is the integer $k$ such that $2^{k-1}\le N&lt;2^k$.

In most implementations (and all implementations conforming to PKCS#1), a private exponent $d$ has size in bits at most the key size $k$, and typically is few bits smaller. Notice that the private exponent is not uniquely defined.

The smallest private exponent is $d=e^{-1}\bmod\lambda(N)$. It is always less than $\lambda(N)$, where $\lambda()$ is Carmichael's function, with $\lambda(N)=\operatorname{lcm}(p-1,q-1)$ when $N=p\cdot q$ with $p$ and $q$ distinct primes. The smallest private exponent always has size in bits strictly less than the key size $k$, and typically is a few bits smaller. 

Sometime, a private exponent is computed as $d=e^{-1}\bmod\varphi(N)$, where $\varphi()$ is Euler's function, with $\varphi(N)=(p-1)\cdot(q-1)$ when $N=p\cdot q$ with $p$ and $q$ distinct primes. That particular $d$ has size in bits at most the key size $k$, often is a few bits smaller, and often is a few bits larger than the smallest private exponent.

Yet other methods to define a private exponent give no maximum for $d$, and only require that $d\equiv e^{-1}\pmod{\lambda(N)}$ (which is the necessary and sufficient condition for $d$ to work), or $d\equiv e^{-1}\pmod{\varphi(N)}$.